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• NCERT Solutions for Class 10 Maths Chapter 4 - Quadratic Equations
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NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations - Free PDF

Maths is a subject that deals with lots of formulas and tricks. It requires a good understanding and lots of practice. Quadratic Equation is a chapter in class 10, which is quite tricky and difficult. But, it has good weightage from the board examination’s point of view. This is the reason why Vedantu has prepared NCERT Class 10 Maths Chapter 4 Solutions. NCERT Solutions of Class 10 Maths Chapter 4 available at Vedantu’s website and app is very simple and easy to understand. Students acquire a complete knowledge about the basic concepts in detail step by step first and then go through the solved examples. NCERT Solutions for Class 10 Science is also available on Vedantu.

Also, students can download Quadratic Equations Class 10 Solutions in PDF form to use offline. Those who want to secure the highest marks in their board examinations should not miss this Maths Chapter 4 Class 10. It has been recorded that every year, at least two or three questions are asked from this chapter. Therefore, it is advised to every student to start practice for Class 10 Maths Ch 4 with NCERT Solutions Class 10 Maths Ch 4 offered by Vedantu.

All Topics of NCERT Class 10 Maths Chapter 4 - Quadratic Equations

The topics covered under Chapter 4 Maths Class 10 are given below.

Important Points

A quadratic equation can be represented as:

ax 2 + bx + c = 0

Where x is the variable of the equation and a, b and c are the real numbers. Also, a≠0.

The nature of roots of a quadratic equation ax 2 + bx + c = 0 can be find as: 

A real number α be root of quadratic equations ax 2 + bx + c = 0 if and only if 

aα 2 + bα + c = 0.

Quadratic equations are very important in real-life situations. Learn all the concepts deeply and understand each topic conceptually. And, now let us solve questions related to quadratic equations.

Related Chapters

Exercises under NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations

Chapter 4 of NCERT Solutions for Class 10 Maths is titled "Quadratic Equations". In this chapter, students will learn about the standard form of a quadratic equation, methods for solving quadratic equations, and the nature of the roots of quadratic equations.

The chapter includes four exercises, each of which covers different aspects of quadratic equations.

Exercise 4.1:

This exercise covers the introduction to quadratic equations and the standard form of a quadratic equation. It also includes methods for solving quadratic equations by factorisation. In this exercise, students will learn how to identify a quadratic equation, how to convert a quadratic equation into standard form, and how to factorise quadratic equations using different methods. The exercise includes a set of questions that range from easy to difficult, allowing students to gradually build their understanding of the concepts.

Exercise 4.2:

This exercise covers more advanced methods for solving quadratic equations, such as completing the square and using the quadratic formula. It includes the derivation of the quadratic formula and shows how to apply it to solve quadratic equations. In this exercise, students will learn how to complete the square of a quadratic equation to convert it into standard form, and how to use the quadratic formula to solve quadratic equations. The exercise includes questions that require students to use both methods to solve quadratic equations.

Exercise 4.3:

This exercise focuses on the nature of the roots of quadratic equations and the discriminant of a quadratic equation. Students will learn how to determine the nature of the roots of a quadratic equation based on the value of its discriminant. The exercise covers the relationship between the coefficients and roots of a quadratic equation, and how to find the sum and product of the roots. The exercise includes a set of questions that require students to apply their knowledge of discriminant and the nature of roots to solve quadratic equations.

Exercise 4.4:

This exercise covers real-life applications of quadratic equations and includes word problems that require students to apply their knowledge of quadratic equations to solve practical problems. The exercise includes problems related to the trajectory of a projectile, finding the distance between two ships, and the dimensions of a garden. Students will learn how to formulate and solve quadratic equations to solve real-life problems. The exercise includes a set of word problems that gradually increase in difficulty, allowing students to develop their problem-solving skills.

Access NCERT Solutions for Class - 10 Maths Chapter 4 – Quadratic Equations

Exercise 4.1

1. Check whether the following are quadratic equations:  

i. ${{\left( \text{x+1} \right)}^{\text{2}}}\text{=2}\left( \text{x-3} \right)$

Ans : ${{\left( \text{x+1} \right)}^{\text{2}}}\text{=2}\left( \text{x-3} \right)$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+2x+1=2x-6}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+7=0}$

Since, it is in the form of $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=0}$.

Therefore, the given equation is a quadratic equation.

ii. ${{\text{x}}^{\text{2}}}\text{-2x=}\left( \text{-2} \right)\left( \text{3-x} \right)$

Ans : ${{\text{x}}^{\text{2}}}\text{-2x=}\left( \text{-2} \right)\left( \text{3-x} \right)$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-2x=-6+2x}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-4x+6=0}$

iii. $\left( \text{x-2} \right)\left( \text{x+1} \right)\text{=}\left( \text{x-1} \right)\left( \text{x+3} \right)$

Ans : $\left( \text{x-2} \right)\left( \text{x+1} \right)\text{=}\left( \text{x-1} \right)\left( \text{x+3} \right)$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-x-2=}{{\text{x}}^{\text{2}}}\text{+2x-3}$

$\Rightarrow \text{3x-1=0}$

Since, it is not in the form of $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=0}$.

Therefore, the given equation is not a quadratic equation.

iv. $\left( \text{x-3} \right)\left( \text{2x+1} \right)\text{=x}\left( \text{x+5} \right)$

Ans : $\left( \text{x-3} \right)\left( \text{2x+1} \right)\text{=x}\left( \text{x+5} \right)$

$\Rightarrow \text{2}{{\text{x}}^{\text{2}}}\text{-5x-3=}{{\text{x}}^{\text{2}}}\text{+5x}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-10x-3=0}$

v. $\left( \text{2x-1} \right)\left( \text{x-3} \right)\text{=}\left( \text{x+5} \right)\left( \text{x-1} \right)$

Ans : $\left( \text{2x-1} \right)\left( \text{x-3} \right)\text{=}\left( \text{x+5} \right)\left( \text{x-1} \right)$

$\Rightarrow \text{2}{{\text{x}}^{\text{2}}}\text{-7x+3=}{{\text{x}}^{\text{2}}}\text{+4x-5}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-11x+8=0}$

vi. ${{\text{x}}^{\text{2}}}\text{+3x+1=}{{\left( \text{x-2} \right)}^{\text{2}}}$

Ans : ${{\text{x}}^{\text{2}}}\text{+3x+1=}{{\left( \text{x-2} \right)}^{\text{2}}}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+3x+1=}{{\text{x}}^{\text{2}}}\text{+4-4x}$

$\Rightarrow \text{7x-3=0}$

vii. ${{\left( \text{x+2} \right)}^{\text{3}}}\text{=2x}\left( {{\text{x}}^{\text{2}}}\text{-1} \right)$

Ans : ${{\left( \text{x+2} \right)}^{\text{3}}}\text{=2x}\left( {{\text{x}}^{\text{2}}}\text{-1} \right)$

$\Rightarrow {{\text{x}}^{\text{3}}}\text{+8+6}{{\text{x}}^{\text{2}}}\text{+12x=2}{{\text{x}}^{\text{3}}}\text{-2x}$

$\Rightarrow {{\text{x}}^{\text{3}}}\text{-14x-6}{{\text{x}}^{\text{2}}}\text{-8=0}$

viii. ${{\text{x}}^{\text{3}}}\text{-4}{{\text{x}}^{\text{2}}}\text{-x+1=}{{\left( \text{x-2} \right)}^{\text{3}}}$

Ans : ${{\text{x}}^{\text{3}}}\text{-4}{{\text{x}}^{\text{2}}}\text{-x+1=}{{\left( \text{x-2} \right)}^{\text{3}}}$ 

$\Rightarrow {{\text{x}}^{\text{3}}}\text{-4}{{\text{x}}^{\text{2}}}\text{-x+1=}{{\text{x}}^{\text{3}}}\text{-8-6}{{\text{x}}^{\text{2}}}\text{+12x}$

$\Rightarrow \text{2}{{\text{x}}^{\text{2}}}\text{-13x+9=0}$

2. Represent the following situations in the form of quadratic equations.

i. The area of a rectangular plot is $\text{528 }{{\text{m}}^{\text{2}}}$. The length of the plot (in metres) is one more than twice its breadth. We need to find the length and breadth of the plot.

Ans : Let the breath of the plot be $\text{x m}$.

Thus, length would be-

$\text{Length=}\left( \text{2x+1} \right)\text{m}$

Hence, Area of rectangle $=$$\text{Length }\!\!\times\!\!\text{ breadth}$

So, $\text{528=x}\left( \text{2x+1} \right)$

$\Rightarrow \text{2}{{\text{x}}^{\text{2}}}\text{+x-528=0}$

ii. The product of two consecutive positive integers is $\text{306}$. We need to find the integers.

Ans : Let the consecutive integers be $\text{x}$ and $\text{x+1}$.

Thus, according to question-

$\text{x}\left( \text{x+1} \right)\text{=306}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+x-306=0}$

iii. Rohan’s mother is $\text{26}$ years older than him. The product of their ages (in years) $\text{3}$ years from now will be $\text{360}$. We would like to find Rohan’s present age.

Ans : Let Rohan’s age be $\text{x}$.

Hence, his mother’s age is $\text{x+26}$ .

Now, after $\text{3 years}$.

Rohan’s age will be $\text{x+3}$.

His mother’s age will be $\text{x+29}$ .

So, according to question-

$\left( \text{x+3} \right)\left( \text{x+29} \right)\text{=360}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+3x+29x+87=360}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+32x-273=0}$

iv. A train travels a distance of $\text{480 km}$ at a uniform speed. If the speed had been $\text{8km/h}$ less, then it would have taken $\text{3}$ hours more to cover the same distance. We need to find the speed of the train.

Ans : Let the speed of train be $\text{x km/h}$.

Thus, time taken to travel $\text{482 km}$ is $\dfrac{\text{480}}{\text{x}}\text{hrs}$.

Now, let the speed of train $\text{=}\left( \text{x-8} \right)\text{km/h}$.

Therefore, time taken to travel $\text{480 km}$ is $\left( \dfrac{\text{480}}{\text{x}}+3 \right)\text{hrs}$.

Hence, $\text{speed }\!\!\times\!\!\text{ time=distance}$

i.e $\left( \text{x-8} \right)\left( \dfrac{\text{480}}{\text{x}}\text{+3} \right)\text{=480}$

$\Rightarrow \text{480+3x-}\dfrac{\text{3840}}{\text{x}}\text{-24=480}$

$\Rightarrow \text{3x-}\dfrac{\text{3840}}{\text{x}}\text{=24}$

$\Rightarrow \text{3}{{\text{x}}^{\text{2}}}\text{-24x-3840=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-8x-1280=0}$

Exercise 4.2

1. Find the roots of the following quadratic equations by factorisation:

i. ${{\text{x}}^{\text{2}}}\text{-3x-10=0}$

Ans : ${{\text{x}}^{\text{2}}}\text{-3x-10=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-5x+2x-10}$

$\Rightarrow \text{x}\left( \text{x-5} \right)\text{+2}\left( \text{x-5} \right)$

$\Rightarrow \left( \text{x-5} \right)\left( \text{x+2} \right)$

Therefore, roots of this equation are –

$\text{x-5=0}$ or $\text{x+2=0}$

i.e $\text{x=5}$ or $\text{x=-2}$

ii. $\text{2}{{\text{x}}^{\text{2}}}\text{+x-6=0}$

Ans : $\text{2}{{\text{x}}^{\text{2}}}\text{+x-6=0}$

$\Rightarrow 2{{\text{x}}^{\text{2}}}\text{+4x-3x-6}$

$\Rightarrow 2\text{x}\left( \text{x+2} \right)-3\left( \text{x+2} \right)$

$\Rightarrow \left( \text{x+2} \right)\left( \text{2x-3} \right)$

$\text{x+2=0}$ or $\text{2x-3=0}$

i.e $\text{x=-2}$ or $\text{x=}\dfrac{3}{2}$

iii. $\sqrt{\text{2}}{{\text{x}}^{\text{2}}}\text{+7x+5}\sqrt{\text{2}}\text{=0}$

Ans : $\sqrt{\text{2}}{{\text{x}}^{\text{2}}}\text{+7x+5}\sqrt{\text{2}}\text{=0}$

$\Rightarrow \sqrt{\text{2}}{{\text{x}}^{\text{2}}}\text{+5x+2x+5}\sqrt{\text{2}}$

$\Rightarrow \text{x}\left( \sqrt{\text{2}}\text{x+5} \right)+\sqrt{\text{2}}\left( \sqrt{\text{2}}\text{x+5} \right)$

$\Rightarrow \left( \sqrt{\text{2}}\text{x+5} \right)\left( \text{x+}\sqrt{\text{2}} \right)$

$\sqrt{\text{2}}\text{x+5=0}$ or $\text{x+}\sqrt{\text{2}}\text{=0}$

i.e $\text{x=}\dfrac{-5}{\sqrt{\text{2}}}$ or $\text{x=-}\sqrt{\text{2}}$

iv. $\text{2}{{\text{x}}^{\text{2}}}\text{-x+}\dfrac{\text{1}}{\text{8}}\text{=0}$

Ans : $\text{2}{{\text{x}}^{\text{2}}}\text{-x+}\dfrac{\text{1}}{\text{8}}\text{=0}$

\[\Rightarrow \dfrac{\text{1}}{\text{8}}\left( 16{{\text{x}}^{\text{2}}}-8x+1 \right)\]

\[\Rightarrow \dfrac{\text{1}}{\text{8}}\left( 4x\left( 4x-1 \right)-1\left( 4x-1 \right) \right)\]

$\Rightarrow \dfrac{\text{1}}{\text{8}} {{\left( \text{4x-1} \right)}^{2}}$

$\text{4x-1=0}$ or $\text{4x-1=0}$

i.e $\text{x=}\dfrac{1}{4}$ or $\text{x=}\dfrac{1}{4}$

v. $\text{100}{{\text{x}}^{\text{2}}}\text{-20x+1=0}$

Ans : $\text{100}{{\text{x}}^{\text{2}}}\text{-20x+1=0}$

$\Rightarrow 100{{\text{x}}^{\text{2}}}\text{-10x-10x+1}$

$\Rightarrow 10\text{x}\left( \text{10x-1} \right)-1\left( \text{10x-1} \right)$

\[\Rightarrow \left( \text{10x-1} \right)\left( \text{10x-1} \right)\]

\[\left( \text{10x-1} \right)=0\]or \[\left( \text{10x-1} \right)=0\]

i.e $\text{x=}\dfrac{1}{10}$ or $\text{x=}\dfrac{1}{10}$

2. i. John and Jivanti together have $\text{45}$ marbles. Both of them lost $\text{5}$ marbles each, and the product of the number of marbles they now have is $\text{124}$. Find out how many marbles they had to start with.

Ans : Let the number of john’s marbles be $\text{x}$.

Thus, number of Jivanti’s marble be $\text{45-x}$.

According to question i.e, 

After losing $\text{5}$ marbles.

Number of john’s marbles be $\text{x-5}$

And number of Jivanti’s marble be $\text{40-x}$.

Therefore, $\left( \text{x-5} \right)\left( \text{40-x} \right)\text{=124}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-45x+324=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-36x-9x+324=0}$

$\Rightarrow \text{x}\left( \text{x-36} \right)\text{-9}\left( \text{x-36} \right)\text{=0}$

$\Rightarrow \left( \text{x-36} \right)\left( \text{x-9} \right)\text{=0}$

Case 1 - If $\text{x-36=0}$ i.e $\text{x=36}$

So, the number of john’s marbles be $\text{36}$.

Thus, number of Jivanti’s marble be $\text{9}$.

Case 2 - If $\text{x-9=0}$ i.e $\text{x=9}$

So, the number of john’s marbles be $9$.

Thus, number of Jivanti’s marble be $36$.

ii. A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be $\text{55}$ minus the number of toys produced in a day. On a particular day, the total cost of production was Rs $\text{750}$. Find out the number of toys produced on that day.

Ans: Let the number of toys produced be $\text{x}$.

Therefore, Cost of production of each toy be $\text{Rs}\left( \text{55-x} \right)$.

Thus, $\left( \text{55-x} \right)\text{x=750}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-55x+750=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-25x-30x+750=0}$

$\Rightarrow \text{x}\left( \text{x-25} \right)-30\left( \text{x-25} \right)\text{=0}$

$\Rightarrow \left( \text{x-25} \right)\left( \text{x-30} \right)\text{=0}$

Case 1 - If $\text{x-25=0}$ i.e $\text{x=25}$

So, the number of toys be $25$.

Case 2 - If $\text{x-30=0}$ i.e $\text{x=30}$

So, the number of toys be $30$.

3. Find two numbers whose sum is $\text{27}$ and product is $\text{182}$ .

Ans: Let the first number be $\text{x}$ ,

Thus, the second number be $\text{27-x}$.

$\text{x}\left( \text{27-x} \right)\text{=182}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-27x+182=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-13x-14x+182=0}$

$\Rightarrow \text{x}\left( \text{x-13} \right)-14\left( \text{x-13} \right)\text{=0}$

$\Rightarrow \left( \text{x-13} \right)\left( \text{x-14} \right)\text{=0}$

Case 1 - If $\text{x-13=0}$ i.e $\text{x=13}$

So, the first number be $13$ ,

Thus, the second number be $\text{14}$.

Case 2 - If $\text{x-14=0}$ i.e $\text{x=14}$

So, the first number be $\text{14}$.

Thus, the second number be$13$.

4. Find two consecutive positive integers, sum of whose squares is $\text{365}$.

Ans: Let the consecutive positive integers be $\text{x}$ and $\text{x+1}$.

Thus, ${{\text{x}}^{\text{2}}}\text{+}{{\left( \text{x+1} \right)}^{\text{2}}}\text{=365}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+}{{\text{x}}^{\text{2}}}+1+2\text{x=365}$

$\Rightarrow 2{{\text{x}}^{\text{2}}}\text{+2x-364=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+x-182=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+14x-13x-182=0}$

$\Rightarrow \text{x}\left( \text{x+14} \right)-13\left( \text{x+14} \right)\text{=0}$

$\Rightarrow \left( \text{x+14} \right)\left( \text{x-13} \right)\text{=0}$

Case 1 - If $\text{x+14=0}$ i.e $\text{x=-14}$.

This case is rejected because number is positive.

Case 2 - If $\text{x-13=0}$ i.e $\text{x=13}$

So, the first number be $\text{13}$.

Thus, the second number be $14$.

Hence, the two consecutive positive integers are $\text{13}$ and $14$.

5. The altitude of a right triangle is $\text{7 cm}$ less than its base. If the hypotenuse is $\text{13 cm}$, find the other two sides.

Ans: Let the base of the right-angled triangle be $\text{x cm}$.

Its altitude be $\left( \text{x-7} \right)\text{cm}$.

Thus, by pythagores theorem-

$\text{bas}{{\text{e}}^{\text{2}}}\text{+altitud}{{\text{e}}^{\text{2}}}\text{=hypotenus}{{\text{e}}^{\text{2}}}$

\[\therefore {{\text{x}}^{\text{2}}}\text{+}{{\left( \text{x-7} \right)}^{\text{2}}}\text{=1}{{\text{3}}^{\text{2}}}\]

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+}{{\text{x}}^{\text{2}}}+49-14\text{x=169}$

$\Rightarrow 2{{\text{x}}^{\text{2}}}\text{-14x-120=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-7x-60=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+12x+5x-60=0}$

$\Rightarrow \text{x}\left( \text{x-12} \right)+5\left( \text{x-12} \right)\text{=0}$

$\Rightarrow \left( \text{x-12} \right)\left( \text{x+5} \right)\text{=0}$

Case 1 - If $\text{x-12=0}$ i.e $\text{x=12}$.

So, the base of the right-angled triangle be $\text{12 cm}$ and Its altitude be $\text{5cm}$

Case 2 - If $\text{x+5=0}$ i.e $\text{x=-5}$

This case is rejected because side is always positive.

Hence, the base of the right-angled triangle be $\text{12 cm}$ and Its altitude be $\text{5cm}$.

6. A cottage industry produces a certain number of pottery articles in a day. It was observed on a particular day that the cost of production of each article (in rupees) was $\text{3}$ more than twice the number of articles produced on that day. If the total cost of production on that day was Rs $\text{90}$, find the number of articles produced and the cost of each article.

Ans:  Let the number of articles produced be $\text{x}$.

Therefore, cost of production of each article be $\text{Rs}\left( \text{2x+3} \right)$.

Thus, $\text{x}\left( \text{2x+3} \right)\text{=90}$

$\Rightarrow 2{{\text{x}}^{\text{2}}}\text{+3x-90=0}$

$\Rightarrow 2{{\text{x}}^{\text{2}}}\text{+15x-12x-90=0}$

$\Rightarrow \text{x}\left( \text{2x+15} \right)-6\left( \text{2x+15} \right)\text{=0}$

$\Rightarrow \left( \text{2x+15} \right)\left( \text{x-6} \right)\text{=0}$

Case 1 - If $\text{2x-15=0}$ i.e $\text{x=}\dfrac{-15}{2}$.

This case is rejected because number of articles is always positive.

Case 2 - If $\text{x-6=0}$ i.e $\text{x=6}$

Hence, the number of articles produced be $6$.

Therefore, cost of production of each article be $\text{Rs15}$.

Exercise 4.3

1. Find the roots of the following quadratic equations, if they exist, by the method of completing the square:

i. $\text{2}{{\text{x}}^{\text{2}}}\text{-7x+3=0}$

Ans: $\text{2}{{\text{x}}^{\text{2}}}\text{-7x+3=0}$

On dividing both sides of the equation by $\text{2}$.

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-}\dfrac{\text{7}}{\text{2}}\text{x=-}\dfrac{\text{3}}{\text{2}}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-2}\left( \dfrac{\text{7}}{\text{4}} \right)\text{x=-}\dfrac{\text{3}}{\text{2}}$

On adding ${{\left( \dfrac{7}{4} \right)}^{2}}$ both sides of equation.

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-2}\left( \dfrac{\text{7}}{\text{4}} \right)\text{x+}{{\left( \dfrac{7}{4} \right)}^{2}}\text{=-}\dfrac{\text{3}}{\text{2}}+{{\left( \dfrac{7}{4} \right)}^{2}}$

$\Rightarrow {{\left( \text{x-}\dfrac{\text{7}}{\text{4}} \right)}^{\text{2}}}\text{=}\dfrac{\text{49}}{\text{16}}\text{-}\dfrac{\text{3}}{\text{2}}$

$\Rightarrow {{\left( \text{x-}\dfrac{\text{7}}{\text{4}} \right)}^{\text{2}}}\text{=}\dfrac{\text{25}}{\text{16}}$

$\Rightarrow \left( \text{x-}\dfrac{\text{7}}{\text{4}} \right)\text{=}\pm \dfrac{5}{4}$

$\Rightarrow x\text{=}\dfrac{\text{7}}{\text{4}}\pm \dfrac{5}{4}$

$\Rightarrow x\text{=}\dfrac{\text{7}}{\text{4}}+\dfrac{5}{4}$ or $x\text{=}\dfrac{\text{7}}{\text{4}}-\dfrac{5}{4}$

$\Rightarrow x\text{=}\dfrac{12}{\text{4}}$ or $x\text{=}\dfrac{2}{\text{4}}$

$\Rightarrow x\text{=}3$ or $x\text{=}\dfrac{\text{1}}{\text{2}}$

ii. $\text{2}{{\text{x}}^{\text{2}}}\text{+x-4=0}$

Ans: $\text{2}{{\text{x}}^{\text{2}}}\text{+x-4=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+}\dfrac{1}{2}\text{x=2}$

On adding ${{\left( \dfrac{1}{4} \right)}^{2}}$ both sides of equation.

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+2}\left( \dfrac{1}{4} \right)\text{x+}{{\left( \dfrac{1}{4} \right)}^{2}}\text{=2+}{{\left( \dfrac{1}{4} \right)}^{2}}$

$\Rightarrow {{\left( \text{x+}\dfrac{\text{1}}{\text{4}} \right)}^{\text{2}}}\text{=}\dfrac{\text{33}}{\text{16}}$

\[\Rightarrow \left( \text{x+}\dfrac{\text{1}}{\text{4}} \right)\text{=}\pm \dfrac{\sqrt{\text{33}}}{4}\]

\[\Rightarrow \text{x= }\!\!\pm\!\!\text{ }\dfrac{\sqrt{\text{33}}}{\text{4}}\text{-}\dfrac{\text{1}}{\text{4}}\]

\[\Rightarrow \text{x= }\!\!\pm\!\!\text{ }\dfrac{\sqrt{\text{33}}-1}{\text{4}}\]

\[\Rightarrow \text{x=}\dfrac{\sqrt{\text{33}}-1}{\text{4}}\] or \[\Rightarrow \text{x=}\dfrac{\sqrt{\text{-33}}-1}{\text{4}}\]

iii. $\text{4}{{\text{x}}^{\text{2}}}\text{+4}\sqrt{3}\text{x+3=0}$

Ans: $\text{4}{{\text{x}}^{\text{2}}}\text{+4}\sqrt{3}\text{x+3=0}$

$\Rightarrow {{\left( 2\text{x} \right)}^{\text{2}}}\text{+2}\left( 2\sqrt{3} \right)\text{x+}{{\left( \sqrt{\text{3}} \right)}^{2}}\text{=0}$

$\Rightarrow {{\left( \text{2x+}\sqrt{\text{3}} \right)}^{\text{2}}}\text{=0}$

$\Rightarrow \left( \text{2x+}\sqrt{\text{3}} \right)\text{=0}$ and $\Rightarrow \left( \text{2x+}\sqrt{\text{3}} \right)\text{=0}$

$\Rightarrow \text{x=}\dfrac{\text{-}\sqrt{\text{3}}}{\text{2}}$ and $\Rightarrow \text{x=}\dfrac{\text{-}\sqrt{\text{3}}}{\text{2}}$

iv. $\text{2}{{\text{x}}^{\text{2}}}\text{+x+4=0}$

Ans: $\text{2}{{\text{x}}^{\text{2}}}\text{+x+4=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+}\dfrac{\text{1}}{\text{2}}\text{x+2=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+2}\left( \dfrac{\text{1}}{4} \right)\text{x=-2}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+2}\left( \dfrac{\text{1}}{4} \right)\text{x+}{{\left( \dfrac{1}{4} \right)}^{2}}\text{=-2+}{{\left( \dfrac{1}{4} \right)}^{2}}$

$\Rightarrow {{\left( \text{x+}\dfrac{\text{1}}{\text{4}} \right)}^{\text{2}}}\text{=}\dfrac{\text{1}}{\text{16}}\text{-2}$

$\Rightarrow {{\left( \text{x+}\dfrac{\text{1}}{\text{4}} \right)}^{\text{2}}}\text{=-}\dfrac{\text{31}}{\text{16}}$

Since, the square of a number cannot be negative.

Therefore, there is no real root for the given equation.

2. Find the roots of the quadratic equations given in Q.1 above by applying the quadratic formula.

Ans : $\text{2}{{\text{x}}^{\text{2}}}\text{-7x+3=0}$

On comparing this equation with $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=}0$.

So, $\text{a=2}$, $\text{b=-7}$, $\text{c=3}$.

Therefore, by using quadratic formula-

$\text{x=}\dfrac{\text{-b }\!\!\pm\!\!\text{ }\sqrt{{{\text{b}}^{\text{2}}}\text{-4ac}}}{\text{2a}}$

$\Rightarrow \text{x=}\dfrac{\text{7 }\!\!\pm\!\!\text{ }\sqrt{\text{49-24}}}{4}$

$\Rightarrow \text{x=}\dfrac{\text{7 }\!\!\pm\!\!\text{ 5}}{4}$

$\Rightarrow \text{x=}\dfrac{\text{7+5}}{4}$  or $\Rightarrow \text{x=}\dfrac{\text{7-5}}{4}$

$\Rightarrow \text{x=}\dfrac{12}{4}$  or $\Rightarrow \text{x=}\dfrac{\text{2}}{4}$

$\therefore \text{x=3 or }\dfrac{\text{1}}{\text{2}}$.

So, $\text{a=2}$, $\text{b=1}$, $\text{c=-4}$.

$\Rightarrow \text{x=}\dfrac{\text{-1 }\!\!\pm\!\!\text{ }\sqrt{\text{1-32}}}{4}$

$\Rightarrow \text{x=}\dfrac{\text{-1 }\!\!\pm\!\!\text{ }\sqrt{33}}{4}$

$\Rightarrow \text{x=}\dfrac{\text{-1+}\sqrt{33}}{4}$  or $\Rightarrow \text{x=}\dfrac{\text{-1-}\sqrt{33}}{4}$

$\therefore \text{x=}\dfrac{\text{-1+}\sqrt{33}}{4}\text{ or }\dfrac{\text{-1-}\sqrt{33}}{4}$.

So, $\text{a=4}$, $\text{b=4}\sqrt{3}$, $\text{c=3}$.

$\Rightarrow \text{x=}\dfrac{\text{-4}\sqrt{3}\text{ }\!\!\pm\!\!\text{ }\sqrt{\text{48-48}}}{8}$

$\Rightarrow \text{x=}\dfrac{\text{-4}\sqrt{3}}{8}$

$\Rightarrow \text{x=}\dfrac{\text{-}\sqrt{3}}{2}$  or $\Rightarrow \text{x=}\dfrac{\text{-}\sqrt{3}}{2}$

$\therefore \text{x=}\dfrac{\text{-}\sqrt{3}}{2}\text{ or }\dfrac{\text{-}\sqrt{3}}{2}$.

So, $\text{a=2}$, $\text{b=1}$, $\text{c=4}$.

\[\Rightarrow \text{x=}\dfrac{\text{-1 }\!\!\pm\!\!\text{ }\sqrt{\text{1-32}}}{4}\]

$\Rightarrow \text{x=}\dfrac{\text{-1 }\!\!\pm\!\!\text{ }\sqrt{\text{-31}}}{4}$

Since, there can not be any negative number inside square root for any real root to exist.

3. Find the roots of the following equations:

i. $\text{x-}\dfrac{\text{1}}{\text{x}}\text{=3,x}\ne \text{0}$

Ans: $\text{x-}\dfrac{\text{1}}{\text{x}}\text{=3,x}\ne \text{0}$

$\Rightarrow\text{1}{{\text{x}}^{\text{2}}}\text{-3x-1=}0$.

So, $\text{a=1}$, $\text{b=-3}$, $\text{c=-1}$.

$\Rightarrow \text{x=}\dfrac{\text{3 }\!\!\pm\!\!\text{ }\sqrt{\text{9+4}}}{2}$

$\Rightarrow \text{x=}\dfrac{\text{3 }\!\!\pm\!\!\text{ }\sqrt{\text{13}}}{2}$

$\Rightarrow \text{x=}\dfrac{\text{3+}\sqrt{\text{13}}}{2}$  or $\Rightarrow \text{x=}\dfrac{\text{3-}\sqrt{\text{13}}}{2}$

$\therefore \text{x=}\dfrac{\text{3+}\sqrt{\text{13}}}{2}\text{ or }\dfrac{\text{3-}\sqrt{\text{13}}}{2}$.

ii. $\dfrac{\text{1}}{\text{x+4}}\text{-}\dfrac{\text{1}}{\text{x-7}}\text{=}\dfrac{\text{11}}{\text{30}}\text{,x}\ne \text{-4,7}$

Ans: $\dfrac{\text{1}}{\text{x+4}}\text{-}\dfrac{\text{1}}{\text{x-7}}\text{=}\dfrac{\text{11}}{\text{30}}\text{,x}\ne \text{-4,7}$

$\Rightarrow \dfrac{\text{x-7-x-4}}{\left( \text{x+4} \right)\left( \text{x-7} \right)}\text{=}\dfrac{\text{11}}{\text{30}}$

$\Rightarrow \dfrac{\text{-11}}{\left( \text{x+4} \right)\left( \text{x-7} \right)}\text{=}\dfrac{\text{11}}{\text{30}}$

$\Rightarrow \left( \text{x+4} \right)\left( \text{x-7} \right)=-30$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-3x-28=-3}0$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-3x+2=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-2x-x+2=0}$

$\Rightarrow \text{x}\left( \text{x-2} \right)\text{-1}\left( \text{x-2} \right)\text{=0}$

$\Rightarrow \left( \text{x-2} \right)\left( \text{x-1} \right)\text{=0}$

$\Rightarrow x\text{=1 or 2}$

4. The sum of the reciprocals of Rehman’s ages, (in years) $\text{3}$ years ago 

and $\text{5}$ years from now is \[\dfrac{\text{1}}{\text{3}}\]. Find his present age.

Ans: Let the present age of Rehman be $\text{x}$ years.

Three years ago, his age was $\left( \text{x-3} \right)\text{years}$.

Five years hence, his age will be $\left( \text{x+5} \right)\text{years}$.

Therefore,$\dfrac{\text{1}}{\text{x-3}}\text{+}\dfrac{\text{1}}{\text{x+5}}\text{=}\dfrac{\text{1}}{\text{3}}$

\[\Rightarrow \dfrac{\text{x+5+x-3}}{\left( \text{x-3} \right)\left( \text{x+5} \right)}\text{=}\dfrac{\text{1}}{\text{3}}\]

\[\Rightarrow \dfrac{\text{2x+2}}{\left( \text{x-3} \right)\left( \text{x+5} \right)}\text{=}\dfrac{\text{1}}{\text{3}}\]

$\Rightarrow 3\left( \text{2x+2} \right)=\left( \text{x-3} \right)\left( \text{x+5} \right)$

$\Rightarrow \text{6x+6=}{{\text{x}}^{\text{2}}}\text{+2x-15}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-4x-21=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-7x+3x-21=0}$

$\Rightarrow \text{x}\left( \text{x-7} \right)\text{+3}\left( \text{x-7} \right)\text{=0}$

$\Rightarrow \left( \text{x-7} \right)\left( \text{x+3} \right)\text{=0}$

$\Rightarrow x\text{=7 or -3}$

Therefore, Rehman’s age is $\text{7 years}$.

5. In a class test, the sum of Shefali’s marks in Mathematics and English is $\text{30}$. Had she got $\text{2}$ marks more in Mathematics and $\text{3}$ marks less in English, the product of their marks would have been $\text{210}$. Find her marks in the two subjects.

Ans: Let the marks in maths be $\text{x}$.

Thus, marks in English will be $\text{30-x}$.

Hence, according to question –

$\left( \text{x+2} \right)\left( \text{30-x-3} \right)\text{=210}$

$\left( \text{x+2} \right)\left( \text{27-x} \right)\text{=210}$

$\Rightarrow \text{-}{{\text{x}}^{\text{2}}}\text{+25x+54=210}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-25x+156=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-12x-13x+156=0}$

$\Rightarrow \text{x}\left( \text{x-12} \right)\text{-13}\left( \text{x-12} \right)\text{=0}$

$\Rightarrow \left( \text{x-12} \right)\left( \text{x-13} \right)\text{=0}$

$\Rightarrow \text{x=12,13}$

Case 1 - If the marks in mathematics are $\text{12}$ , then marks in English will be $18$.

Case 2 - If the marks in mathematics are $\text{13}$ , then marks in English will be $17$.

6. The diagonal of a rectangular field is $\text{60}$ metres more than the shorter side. If the longer side is $\text{30}$ metres more than the shorter side, find the sides of the field.

Ans: Let the shorter side of the rectangle be $\text{x m}$.

Thus, Larger side of the rectangle will be $\left( \text{x+30} \right)\text{m}$.

Diagonal of the rectangle be $\sqrt{{{\text{x}}^{\text{2}}}\text{+}{{\left( \text{x+30} \right)}^{\text{2}}}}$

Hence, according to question-

$\sqrt{{{\text{x}}^{\text{2}}}\text{+}{{\left( \text{x+30} \right)}^{\text{2}}}}\text{=x+60}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+}{{\left( \text{x+30} \right)}^{\text{2}}}\text{=}{{\left( \text{x+60} \right)}^{2}}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+}{{\text{x}}^{\text{2}}}\text{+900+60x=}{{\text{x}}^{\text{2}}}\text{+3600+120x}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-60x-2700=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-90x+30x-2700=0}$

$\Rightarrow \text{x}\left( \text{x-90} \right)+30\left( \text{x-90} \right)\text{=0}$

$\Rightarrow \left( \text{x-90} \right)\left( \text{x+30} \right)\text{=0}$

$\Rightarrow \text{x=90,-30}$

Since, side cannot be negative. 

Therefore, the length of the shorter side of rectangle is $\text{90 m}$.

Hence, length of the larger side of the rectangle be $\text{120 m}$.

7. The difference of squares of two numbers is $\text{180}$. The square of the smaller number is $\text{8}$ times the larger number. Find the two numbers.

Ans: Let the larger number be $\text{x}$ and smaller number be $\text{y}$.

According to question-

${{\text{x}}^{\text{2}}}\text{-}{{\text{y}}^{\text{2}}}\text{=180}$ and ${{\text{y}}^{\text{2}}}\text{=8x}$

\[\Rightarrow {{\text{x}}^{\text{2}}}\text{-8x=180}\]

\[\Rightarrow {{\text{x}}^{\text{2}}}\text{-8x-180=0}\]

\[\Rightarrow {{\text{x}}^{\text{2}}}\text{-18x+10x-180=0}\]

$\Rightarrow \text{x}\left( \text{x-18} \right)+10\left( \text{x-18} \right)\text{=0}$

$\Rightarrow \left( \text{x-18} \right)\left( \text{x+10} \right)\text{=0}$

$\Rightarrow \text{x=18,-10}$

Since, larger cannot be negative as $8$ times of the larger number will be negative and hence, the square of the smaller number will be negative which is not possible.

Therefore, the larger number will be $18$.

$\therefore {{\text{y}}^{\text{2}}}\text{=8}\left( \text{18} \right)$

$\Rightarrow {{\text{y}}^{\text{2}}}\text{=144}$

$\Rightarrow \text{y= }\!\!\pm\!\!\text{ 12}$

Hence, smaller number be $\pm 12$.

Therefore, the numbers are $18$ and $12$ or $18$ and $-12$ .

8. A train travels $\text{360 km}$km at a uniform speed. If the speed had been $\text{5km/h}$ more, it would have taken $\text{1}$hour less for the same journey. Find the speed of the train.

Ans: Let the speed of the train be $\text{x km/h}$.

Time taken to cover $\text{360 km/h}$ be $\dfrac{\text{360}}{\text{x}}$.

$\left( \text{x+5} \right)\left( \dfrac{\text{360}}{\text{x}}\text{-1} \right)\text{=360}$

$\Rightarrow \text{360-x+}\dfrac{\text{1800}}{\text{x}}\text{-5=360}$ 

\[\Rightarrow {{\text{x}}^{\text{2}}}\text{+5x-1800=0}\]

\[\Rightarrow {{\text{x}}^{\text{2}}}\text{+45x-40x-1800=0}\]

$\Rightarrow \text{x}\left( \text{x+45} \right)-40\left( \text{x+45} \right)\text{=0}$

$\Rightarrow \left( \text{x+45} \right)\left( \text{x-40} \right)\text{=0}$

$\Rightarrow \text{x=40,-45}$

Since, the speed cannot be negative.

Therefore, the speed of the train is $\text{40 km/h}$.

9. Two water taps together can fill a tank in $\text{9}\dfrac{\text{3}}{\text{8}}$ hours. The tap of larger diameter takes $\text{10}$ hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank.

Ans: Let the time taken by the smaller pipe to fill the tank be $\text{x hr}$.

So, time taken by larger pipe be $\left( \text{x-10} \right)\text{hr}$.

Part of the tank filled by smaller pipe in $1$ hour is $\dfrac{\text{1}}{\text{x}}$.

Part of the tank filled by larger pipe in $1$ hour is $\dfrac{\text{1}}{\text{x-10}}$.

So, according to the question-

$\dfrac{\text{1}}{\text{x}}\text{+}\dfrac{\text{1}}{\text{x-10}}\text{=9}\dfrac{\text{3}}{\text{8}}$

$\Rightarrow \dfrac{\text{1}}{\text{x}}\text{+}\dfrac{\text{1}}{\text{x-10}}\text{=}\dfrac{\text{75}}{\text{8}}$

\[\Rightarrow \dfrac{\text{x-10+x}}{\text{x}\left( \text{x-10} \right)}\text{=}\dfrac{\text{8}}{\text{75}}\]

\[\Rightarrow \dfrac{\text{2x-10}}{\text{x}\left( \text{x-10} \right)}\text{=}\dfrac{\text{8}}{\text{75}}\]

$\Rightarrow \text{75}\left( \text{2x-10} \right)\text{=8}{{\text{x}}^{\text{2}}}\text{-80x}$

$\Rightarrow \text{150x-750=8}{{\text{x}}^{\text{2}}}\text{-80x}$

$\Rightarrow \text{8}{{\text{x}}^{\text{2}}}\text{-230x+750=0}$

$\Rightarrow \text{8}{{\text{x}}^{\text{2}}}\text{-200x-30x+750=0}$

$\Rightarrow \text{8x}\left( \text{x-25} \right)\text{-30}\left( \text{x-25} \right)\text{=0}$

$\Rightarrow \left( \text{x-25} \right)\left( \text{8x-30} \right)\text{=0}$

$\Rightarrow x\text{=25 or }\dfrac{\text{30}}{\text{8}}$

Case 1- If time taken by smaller pipe be $\dfrac{\text{30}}{\text{8}}$ i.e $\text{3}\text{.75 hours}$.

So, Time taken by larger pipe will be negative which is not possible.

Hence, this case is rejected.

Case 2- If the time taken by smaller pipe be $\text{25}$.Then, time taken by larger pipe will be $\text{15 hours}$.

Therefore, time taken by smaller pipe be $\text{25 hours}$ and time taken by larger pipe will be $\text{15 hours}$.

10. An express train takes $\text{1}$ hour less than a passenger train to travel $\text{132 km}$ between Mysore and Bangalore (without taking into consideration the time they stop at intermediate stations). If the average speeds of the express train is $\text{11 km/h}$ more than that of the passenger train, find the average speed of the two trains.

Ans: Let the average speed of passenger train be $\text{x km/h}$.

So, Average speed of express train be $\left( \text{x+11} \right)\text{km/h}$.

Thus, according to question.

$\therefore \dfrac{\text{132}}{\text{x}}\text{-}\dfrac{\text{132}}{\text{x+11}}\text{=1}$

 $\Rightarrow \text{132}\left[ \dfrac{\text{x+11-x}}{\text{x}\left( \text{x+11} \right)} \right]\text{=1}$

$\Rightarrow \dfrac{\text{132 }\!\!\times\!\!\text{ 11}}{\text{x}\left( \text{x+11} \right)}\text{=1}$

$\Rightarrow \text{132 }\!\!\times\!\!\text{ 11=x}\left( \text{x+11} \right)$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+11x-1452=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{+44x-33x-1452=0}$

$\Rightarrow \text{x}\left( \text{x+44} \right)\text{-33}\left( \text{x+44} \right)\text{=0}$

$\Rightarrow \left( \text{x+44} \right)\left( \text{x-33} \right)\text{=0}$

$\Rightarrow x\text{=-44 or 33}$

Since, speed cannot be negative.

Therefore, the speed of the passenger train will be $\text{33 km/h}$ and thus, the speed of the express train will be $\text{44 km/h}$ .

11. Sum of the areas of two squares is $\text{468 }{{\text{m}}^{\text{2}}}$. If the difference of their perimeters are $\text{24 m}$, find the sides of the two squares .

Ans: Let the sides of the two squares be $\text{x m}$ and $\text{y m}$.

Thus, their perimeters will be $\text{4x}$ and $\text{4y}$ and areas will be ${{\text{x}}^{2}}$ and ${{\text{y}}^{2}}$.

$\text{4x-4y=24}$

$\Rightarrow \text{x-y=6}$

$\Rightarrow \text{x=y+6}$

And ${{\text{x}}^{\text{2}}}\text{+}{{\text{y}}^{\text{2}}}\text{=468}$

Substituting value of x-

${{\left( \text{y+6} \right)}^{\text{2}}}\text{+}{{\text{y}}^{\text{2}}}\text{=468}$

$\Rightarrow \text{36+}{{\text{y}}^{\text{2}}}\text{+12y+}{{\text{y}}^{\text{2}}}\text{=468}$

$\Rightarrow \text{2}{{\text{y}}^{\text{2}}}\text{+12y-432=0}$

$\Rightarrow {{\text{y}}^{\text{2}}}\text{+6y-216=0}$

$\Rightarrow {{\text{y}}^{\text{2}}}\text{+18y-12y-216=0}$

$\Rightarrow \text{y}\left( \text{y+18} \right)\text{-12}\left( \text{y+18} \right)\text{=0}$

$\Rightarrow \left( \text{y+18} \right)\left( \text{y-12} \right)\text{=0}$

$\Rightarrow \text{y=-18 or 12}$

Since, side cannot be negative.

Therefore, the sides of the square are $\text{12 m}$ and $\left( \text{12+6} \right)\text{m}$ i.e $\text{18 m}$.

Exercise 4.4

1. Find the nature of the roots of the following quadratic equations. 

If the real roots exist, find them-

i. $\text{2}{{\text{x}}^{\text{2}}}\text{-3x+5=0}$

Ans: For a quadratic equation $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=0}$.

Where Discriminant $\text{=}{{\text{b}}^{\text{2}}}\text{-4ac}$

Case 1- If ${{\text{b}}^{\text{2}}}\text{-4ac>0}$ then there will be two distinct real roots.

Case 2- If ${{\text{b}}^{\text{2}}}\text{-4ac=0}$ then there will be two equal real roots.

Case 3- If ${{\text{b}}^{\text{2}}}\text{-4ac<0}$ then there will be no real roots.

Thus, for $\text{2}{{\text{x}}^{\text{2}}}\text{-3x+5=0}$ .

So, $\text{a=2}$, $\text{b=-3}$, $\text{c=5}$.

Discriminant $\text{=}{{\left( \text{-3} \right)}^{\text{2}}}\text{-4}\left( \text{2} \right)\left( \text{5} \right)$

$\text{=9-40}$

$\text{=-31}$

Since, Discriminant: ${{\text{b}}^{\text{2}}}\text{-4ac < 0}$.

ii. $\text{3}{{\text{x}}^{\text{2}}}\text{-4}\sqrt{\text{3}}\text{x+4=0}$

Case 1- If ${{\text{b}}^{\text{2}}}\text{-4ac > 0}$ then there will be two distinct real roots.

Case 3- If ${{\text{b}}^{\text{2}}}\text{-4ac < 0}$ then there will be no real roots.

Thus, for $\text{3}{{\text{x}}^{\text{2}}}\text{-4}\sqrt{\text{3}}\text{x+4=0}$ .

So, $\text{a=3}$, $\text{b=-4}\sqrt{\text{3}}$, $\text{c=4}$.

Discriminant $\text{=}{{\left( \text{-4}\sqrt{\text{3}} \right)}^{\text{2}}}\text{-4}\left( \text{3} \right)\left( \text{4} \right)$

$\text{=48-48}$

$\text{=0}$

Since, Discriminant: ${{\text{b}}^{\text{2}}}\text{-4ac=0}$.

Therefore, there is equal real root for the given equation and the roots are-

$\dfrac{\text{-b}}{\text{2a}}$ and $\dfrac{\text{-b}}{\text{2a}}$.

Hence, roots are-

$\dfrac{\text{-b}}{\text{2a}}\text{=}\dfrac{\text{-}\left( \text{-4}\sqrt{\text{3}} \right)}{\text{6}}$

$\text{=}\dfrac{\text{4}\sqrt{\text{3}}}{\text{6}}$

\[\text{=}\dfrac{\text{2}\sqrt{\text{3}}}{3}\]

Therefore, roots are \[\dfrac{\text{2}\sqrt{\text{3}}}{3}\] and \[\dfrac{\text{2}\sqrt{\text{3}}}{3}\].

iii. $\text{2}{{\text{x}}^{\text{2}}}\text{-6x+3=0}$

Thus, for $\text{2}{{\text{x}}^{\text{2}}}\text{-6x+3=0}$ .

So, $\text{a=2}$, $\text{b=-6}$, $\text{c=3}$.

Discriminant $\text{=}{{\left( \text{-6} \right)}^{\text{2}}}\text{-4}\left( \text{2} \right)\left( \text{3} \right)$

$\text{=36-24}$

$\text{=12}$

Since, Discriminant: ${{\text{b}}^{\text{2}}}\text{-4ac>0}$.

Therefore, distinct real roots exists for the given equation and the roots are-

$\text{x=}\dfrac{\text{-}\left( \text{-6} \right)\text{ }\!\!\pm\!\!\text{ }\sqrt{{{\left( \text{-6} \right)}^{\text{2}}}\text{-4}\left( \text{2} \right)\left( \text{3} \right)}}{\text{4}}$

$\text{=}\dfrac{\text{6 }\!\!\pm\!\!\text{ }\sqrt{\text{36-24}}}{\text{4}}$

$\text{=}\dfrac{\text{6 }\!\!\pm\!\!\text{ }\sqrt{\text{12}}}{\text{4}}$

$\text{=}\dfrac{\text{6 }\!\!\pm\!\!\text{ 2}\sqrt{\text{3}}}{\text{4}}$

$\text{=}\dfrac{\text{3 }\!\!\pm\!\!\text{ }\sqrt{\text{3}}}{\text{2}}$

Therefore, roots are $\dfrac{\text{3+}\sqrt{\text{3}}}{\text{2}}$ and $\dfrac{\text{3-}\sqrt{\text{3}}}{\text{2}}$.

2. Find the values of $\text{k}$ for each of the following quadratic equations, so 

that they have two equal roots.

i. $\text{2}{{\text{x}}^{\text{2}}}\text{+kx+3=0}$

Ans: If a quadratic equation $\text{a}{{\text{x}}^{\text{2}}}\text{+bx+c=0}$ has two equal roots, then its discriminant will be $\text{0}$ i.e., ${{\text{b}}^{\text{2}}}\text{-4ac=0}$

So, for $\text{2}{{\text{x}}^{\text{2}}}\text{+kx+3=0}$ .

So, $\text{a=2}$, $\text{b=k}$, $\text{c=3}$.

Discriminant $\text{=}{{\left( \text{k} \right)}^{\text{2}}}\text{-4}\left( \text{2} \right)\left( \text{3} \right)$

$\text{=}{{\text{k}}^{2}}-24$

For equal roots-

${{\text{b}}^{\text{2}}}\text{-4ac=0}$

$\therefore {{\text{k}}^{\text{2}}}\text{-24=0}$

$\Rightarrow {{\text{k}}^{\text{2}}}\text{=24}$

$\Rightarrow \text{k=}\sqrt{\text{24}}$

$\Rightarrow \text{k=}\pm \text{2}\sqrt{\text{6}}$

ii. $\text{kx}\left( \text{x-2} \right)\text{+6=0}$

So, for $\text{kx}\left( \text{x-2} \right)\text{+6=0}$

$\Rightarrow \text{k}{{\text{x}}^{\text{2}}}\text{-2kx+6=0}$

So, $\text{a=k}$, $\text{b=-2k}$, $\text{c=6}$.

Discriminant $\text{=}{{\left( \text{-2k} \right)}^{\text{2}}}\text{-4}\left( \text{k} \right)\left( \text{6} \right)$

$\text{=4}{{\text{k}}^{\text{2}}}\text{-24k}$

$\therefore \text{4}{{\text{k}}^{\text{2}}}\text{-24k=0}$

$\Rightarrow \text{4k}\left( \text{k-6} \right)\text{=0}$

$\Rightarrow \text{k=0 or k=6}$

But $\text{k}$ cannot be zero. Thus, this equation has two equal roots when $\text{k}$ should be $\text{6}$ .

3. Is it possible to design a rectangular mango grove whose length is 

twice its breadth, and the area is $\text{800}{{\text{m}}^{\text{2}}}$ ? If so, find its length and breadth.

Ans: Let the breadth of mango grove be $\text{x}$.

So, length of mango grove will be $\text{2x}$.

Hence, Area of mango grove is $=\left( \text{2x} \right)\text{x}$

$\text{=2}{{\text{x}}^{\text{2}}}$.

So, $\text{2}{{\text{x}}^{\text{2}}}\text{=800}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{=400}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-400=0}$

So, $\text{a=1}$, $\text{b=0}$, $\text{c=400}$.

Discriminant $\text{=}{{\left( \text{0} \right)}^{\text{2}}}\text{-4}\left( \text{1} \right)\left( \text{-400} \right)$

$\text{=1600}$

Therefore, distinct real roots exist for the given equation and the roots are-

$\text{x=}\dfrac{\text{-}\left( 0 \right)\text{ }\!\!\pm\!\!\text{ }\sqrt{{{\left( 0 \right)}^{\text{2}}}\text{-4}\left( 1 \right)\left( -400 \right)}}{2}$

$\text{=}\dfrac{\pm \sqrt{\text{1600}}}{2}$

$\text{=}\dfrac{\text{ }\!\!\pm\!\!\text{ 40}}{2}$

$\text{=}\pm \text{20}$

Since, length cannot be negative.

Therefore, breadth of the mango grove is $\text{20m}$.

And length of the mango grove be $\text{2}\left( \text{20} \right)\text{m}$ i.e., $\text{40m}$.

4. Is the following situation possible? If so, determine their present ages. The sum of the ages of two friends is $\text{20}$ years. Four years ago, the product of their ages in years was $\text{48}$.

Ans: Let the age of one friend be $\text{x years}$.

So, age of the other friend will be $\left( \text{20-x} \right)\text{years}$.

Thus, four years ago, the age of one friend be $\left( \text{x-4} \right)\text{years}$.

And age of the other friend will be $\left( \text{16-x} \right)\text{years}$.

$\left( \text{x-4} \right)\left( \text{16-x} \right)\text{=48}$

$\Rightarrow \text{16x-64-}{{\text{x}}^{\text{2}}}\text{+4x=48}$

$\Rightarrow 20\text{x-112-}{{\text{x}}^{\text{2}}}\text{=0}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-20x+112-=0}$

So, $\text{a=1}$, $\text{b=-20}$, $\text{c=112}$.

Discriminant $\text{=}{{\left( \text{-20} \right)}^{\text{2}}}\text{-4}\left( \text{1} \right)\left( \text{112} \right)$

$\text{=400-448}$

$\text{=-48}$

Since, Discriminant: ${{\text{b}}^{\text{2}}}\text{-4ac <0}$.

Therefore, there is no real root for the given equation and hence, this situation is not possible.

5. Is it possible to design a rectangular park of perimeter $\text{80 m}$ and area $\text{400}{{\text{m}}^{\text{2}}}$? If so find its length and breadth.

Ans: Let the length of the park be $\text{x m}$ and breadth of the park be $\text{x m}$.

Thus, $\text{Perimeter=2}\left( \text{x+y} \right)$.

$\text{2}\left( \text{x+y} \right)\text{=80}$

$\Rightarrow \text{x+y=40}$

$\Rightarrow \text{y=40-x}$.

Now, $\text{Area=x }\!\!\times\!\!\text{ y}$.

Substituting value of y.

$\text{Area=x}\left( \text{40-x} \right)$

$\text{x}\left( \text{40-x} \right)\text{=400}$

$\Rightarrow {{\text{x}}^{\text{2}}}\text{-40x+400=0}$

So, $\text{a=1}$, $\text{b=-40}$, $\text{c=400}$.

Discriminant $\text{=}{{\left( \text{-40} \right)}^{\text{2}}}\text{-4}\left( \text{1} \right)\left( 400 \right)$

$\text{=1600-1600}$

Therefore, there is equal real roots for the given equation and hence, this situation is possible.

$\dfrac{\text{-b}}{\text{2a}}\text{=}\dfrac{\text{-}\left( -40 \right)}{2}$

$\text{=}\dfrac{\text{40}}{2}$

\[\text{=20}\]

Therefore, length of park is $\text{x=20m}$ .

And breadth of park be $\text{y=}\left( \text{40-20} \right)\text{m}$ i.e., $\text{y=20m}$.

Overview of the Exercises Covered in NCERT Solutions for Class 10 Chapter 4 Quadratic Equations

Ex 4.1: There are 2 sums with a total of 12 sub-parts in the NCERT Solutions for Class 10 Maths Ch-4 Exercise 4.1. Students will have to verify if the given equations are quadratic equations or not, for the first few sums. For the sums covered in the latter part of the exercise, students will have to form quadratic equations from the given word problems. These sums will familiarize them with the standard quadratic equation formula ax 2 +bx+c=0 .  

Ex 4.2: There are a total of 6 sums covered in the NCERT Solutions for Class 10 Maths Ch-4 Exercise 4.2. These sums are based on the application of the concept of factorisation in quadratic equations. Students will have to form the quadratic equation from the given word problems and find the roots of the quadratic equations by the middle term factor method.

Ex 4.3: There are a total of 11 sums in the NCERT Solutions for Class 10 Maths Ch-4 Exercise 4.3. Students need to have a good understanding of the concepts of speed, time, and distance, time and work, average speed, perimeter and area, etc. to solve the sums covered in this exercise. The first few sums of this exercise require students to apply the method of completing the perfect square terms in quadratic equations. Also, the some of the sums covered in this exercise have the application of the quadratic formula, 

\[ x = \frac{-b\pm \sqrt{b^{2}-4ac}}{2a} \]

Ex.4.4: There are a total of 5 sums in the NCERT Solutions for Class 10 Maths Ch-4 Exercise 4.4. The sums covered in this exercise will familiarize themselves with the nature of roots i.e., real, imaginary, equal, and unequal roots. The sums will help them to identify and deduce the discriminants of quadratic equations and solve the sums to find the nature of roots.

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations - PDF Download

You can opt for Chapter 4 - Quadratic Equation NCERT Solutions for Class 10 Maths PDF for Upcoming Exams and also You can Find the Solutions of All the Maths Chapters below.

Chapter 1 - Real Numbers

Chapter 2 - Polynomials

Chapter 3 - Pair of Linear Equations in Two Variable

Chapter 5 - Arithmetic Progressions

Chapter 6 - Triangles

Chapter 7 - Coordinate Geometry

Chapter 8 - Introduction to Trigonometry

Chapter 9 - Some Applications of Trigonometry

Chapter 10 - Circles

Chapter 11 - Constructions

Chapter 12 - Areas Related to Circles

Chapter 13 - Surface Areas and Volumes

Chapter 14 - Statistics

Chapter 15 - Probability

Quadratic Equations Class 10 NCERT Solutions have composed in such a way that every student can understand all concepts easily. The expert team of Vedantu has put all their possible efforts in Class 10 CBSE Maths Chapter 4 Solutions to make it exciting and fun-loving. No doubt that Maths is a subject which is not easy to learn. The formulas are not easy to learn, and students don't catch tricks easily where they should apply.

Chapter 4 Maths Class 10 contains quadratic equations to find the value of x. Apart from this, there are approximately three methods given in Class 10 Maths Chapter 4 for this. But, not all three methods are easy to understand to an individual. But in Chapter 4 Maths Class 10 NCERT Solutions, the experts of Vedantu have explained all three methods in very interesting ways that any student can learn quickly. All concepts of quadratic equations and formulas have been explained well. From where the formula comes, how it is discovered, by whom it is discovered, and many other things are mentioned first for the basic knowledge.

After that, many solved examples have been given in NCERT Solution Class 10 Maths Chapter 4 to teach students how to apply formulas and solve numerical problems. All NCERT exercises have been solved by the experts of Vedantu for the students. Some unsolved questions are also included in Class 10 Maths Chapter 4 NCERT Solutions for the practice of the students. Along with this, previous year's questions have also been stated in between the NCERT Solutions. For the ease of the students, it is also mentioned which question came in which year. Some of the mock test papers are also available at the end of pdf for the students' better preparation.  Aspirants who don't want to miss any single question in the board exam should practice regularly with Class 10 NCERT Maths Chapter 4 Solutions.

We can say with full confidence that Vedantu’s NCERT Maths Solution Class 10 Chapter 4 is the single key to get success in board examinations. Students can download the free PDF by a single click on the link available in the website of Vedantu.

Key Features for NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations

Class 10 Maths NCERT Solutions for Chapter 4 on Vedantu are highly recommended by the CBSE 10th Board teachers. These solutions are among the best guides for self-study purposes for the students of Class 10. The key features of these solutions for Quadratic Equations given below explain how these solutions are a must-have for CBSE Term 2 Class 10 Maths exam.

These NCERT Solutions PDFs for Class 10 Quadratic Equations are easy to access online and free to download for offline practice purposes.

Since these NCERT Solution PDFs are available on the Vedantu mobile app, students can download them on their mobile phones and study at any time.

Each sum covered in the four exercises of Class 10 Ch-4 Quadratic Equations is solved in a step-by-step manner for a thorough understanding of all students. 

Students can refer to these NCERT Solutions for their exam preparation as our highly experienced Maths teachers have prepared these solutions as per the latest CBSE guidelines for Class 10.

Students can refer to these NCERT Solutions whenever they are stuck with any sum. Thus, they can address their doubts in Quadratic Equations during the last-minute revisions without wasting any time or having to wait to consult any friend.

Along with this, students can also download additional study materials provided by Vedantu, for Chapter 4 of CBSE Class 10 Maths Solutions –

Chapter 4: Important Questions

Chapter 4: Important Formulas

Chapter 4: Revision Notes

Chapter 4: NCERT Exemplar Solutions

Chapter 4: RD Sharma Solutions

What is the Advantage of Class 10 Maths NCERT Solutions Chapter 4 Provided by Vedantu?

Maths NCERT Solutions Class 10 Chapter 4 includes all solutions for the numerical problem given in the NCERT book. It follows the latest instructions and guidelines announced by the NCERT. Vedantu doesn’t leave any question or concept that is important for the exams. All numerical problems are framed well and solved accurately in NCERT Class 10 Maths Chapter 4 solution. It would not be wrong if we say that the study materials provided by Vedantu is the bible for all those students who want to score good marks.

The subject experts at Vedantu have prepared the NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations for the Term 2 exam preparation and revision purposes for students. To solve the sums covered in this chapter, students should have a sound understanding of other concepts of Maths, like perimeter and area, time and work, linear equations, etc. Therefore, downloading and referring to the Maths NCERT Solutions Class 10 Ch-4 will help students to prepare very well for their exams.

FAQs on NCERT Solutions for Class 10 Maths Chapter 4 - Quadratic Equations

The subject’s experts take online classes available at Vedantu. They have lots of experience in their respective subjects. Along with this, they are working in their field with consistency. So, they have been faced with all kinds of problems and learned the tricks on how to come out from it.

In online classes, teachers share all their experiences with the students and teach some essential tricks. These tricks will be useful at the time of solving questions in the examination hall. Apart from the teaching concepts and formulas, they also motivate students and remove the fear of examinations from the student’s mind that is built up somewhere due to some kind of society’s pressure and other things.

2. Instead of Class 10th Maths Chapter 4, What is the Maths Syllabus for Class 10 for 202 4-25 ?

Here is the complete syllabus for Class 10 Mathematics Revised Syllabus 2024-25:-

Unit- I: Number Systems

Real Numbers

Unit II: Algebra

Polynomials

Pair of Linear Equations in Two Variables

Quadratic Equations

 Arithmetic Progressions

Unit III: Coordinate Geometry

Lines (In two-dimensions)

 Unit IV: Geomtry

Constructions

Unit V: Trigonometry

Introduction to Trigonometry

Trigonometric Identities

 Heights and Distances: Angle of elevation, Angle of Depression

Unit VI: Mensuration

Areas Related to Circles

Surface Areas and Volumes

Unit VII: Statistics and Probability

 Statistics

Probability

3. What is the Weightage for Class 10 Mathematics Unit-Wise?

Students who don’t know the weightage then they should go through the table given below. Here, we have mentioned the entire Class 10 Mathematics Unit-Wise Weightage for the knowledge of the students.

4. Mention the important concepts that you learn in NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations.

If you want to score 100 per cent marks in Class 10 Maths, you have to practice daily. Chapter 4 Quadratic Equations is an important chapter. Students can find NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations on Vedantu. There are five exercises in Chapter 4. Students can download the NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations to learn the important concepts that will help them understand the topic.

5. How to download Class 10 Maths Quadratic Equations NCERT Textbooks PDF?

Students can easily download Class 10 Maths Quadratic Equations NCERT textbooks PDF online. NCERT Solutions for Class 10 Maths Quadratic Equations are explained in an easy and simple language. Follow the given steps:

Click NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations.

Click on “Download PDF”.

Download and save it.

Students can use the PDF document without having an internet connection and can study Maths Quadratic Equations anytime. The NCERT Solutions give a clear understanding to them. 

6. What is the Quadratic formula Class 10th?

When a quadratic polynomial is equated to a constant, it forms a quadratic equation. An equation such as Ax = D, where Ax is a polynomial of degree two and D is a constant, forms a quadratic equation. The standard quadratic equation is a x 2 +bx+c=0 where a, b, and c are not equal to zero. You can refer to NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations to understand more about the topic. You can also download Vedantu’s app. All the resources are available free of cost. 

7. What are the important topics covered in NCERT Solutions Class 10 Maths Chapter 4?

Chapter 4 Class 10 Maths is based on Quadratic Equations. The chapter includes important concepts about quadratic equations. This chapter includes five exercises that explain the different concepts of quadratic equations. The following topics are covered:

Exercise 4.1- Introduction

Exercise 4.2- Quadratic Equations

Exercise 4.3- Solution of a Quadratic Equation by Factorisation

Exercise 4.4- Solution of a Quadratic Equation by Completing the Square

Exercise 4.5- Nature of Roots

8. How do you solve Quadratic Equations in Class 10?

If you want to learn how to solve quadratic equations in Class 10, you can refer to NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations. All the solutions are prepared by experts in an easy language. Students can understand the equations clearly. Students have to find the roots by using the quadratic formula. They can find the sum and product of both the roots. The method is simple and explained properly for easier understanding.

NCERT Solutions for Class 10 Maths

Ncert solutions for class 10.

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations

Revised NCERT Solutions for class 10 Maths Chapter 4 Quadratic Equations in Hindi and English Medium updated for board exams 2024. The question answers and explanation of chapter 4 of 10th Maths are based on NCERT textbooks published for 2024-25. Class 10 Maths Chapter 4 Solutions for CBSE Board Class 10 Maths Chapter 4 Exercise 4.1 Class 10 Maths Chapter 4 Exercise 4.2 Class 10 Maths Chapter 4 Exercise 4.3 Class 10 Maths Chapter 4 Important Questions

Class 10 Maths Chapter 4 Solutions for State Boards Class 10 Maths Exercise 4.1 Class 10 Maths Exercise 4.2 Class 10 Maths Exercise 4.3 Class 10 Maths Exercise 4.4

Get the free Hindi Medium solutions for the academic session 2024-25. Download here UP Board Solutions for Class 10 Maths Chapter 4 all exercises. 10th Maths Chapter 4 solutions are online or download in PDF format. Download Assignments for practice with Solutions 10th Maths Chapter 4 Assignment 1 10th Maths Chapter 4 Assignment 2 10th Maths Chapter 4 Assignment 3 10th Maths Chapter 4 Assignment 4

10th Maths Chapter 4 NCERT Solutions follows the latest CBSE syllabus. Students of MP, UP, Gujarat board and CBSE can use it for Board exams. Class 10 Maths NCERT Solutions Offline Apps in Hindi or English for offline use. For any scholarly help, you may contact us. We will try to help you in the best possible ways.

NCERT Solutions for class 10 Maths Chapter 4 are given below in PDF format or view online. Solutions are in Hindi and English Medium. Uttar Pradesh students also can download UP Board Solutions for Class 10 Maths Chapter 4 here in Hindi Medium.

It is very essential to learn quadratic equations, because it have wide applications in other branches of mathematics, physics, in other subjects and also in real life situations. Download NCERT books 2024-25, revision books and solutions from the links given below.

Previous Year’s CBSE Questions

1. Two marks questions Find the roots of the quadratic equation √2 x² + 7x + 5√2 = 0. [CBSE 2017] 2. Find the value of k for which the equation x²+k(2x + k – 1) + 2 = 0 has real and equal roots. [CBSE 2017] 2. Three marks questions If the equation (1 + m² ) x² +2mcx + c² – a² = 0 has equal roots then show that c² = a² (1 + m²). 3. Four marks questions Speed of a boat in still water is 15 km/h. It goes 30 km upstream and returns back at the same point in 4 hours 30 minutes. Find the speed of the stream. [CBSE 2017]

The word quadratic is derived from the Latin word “Quadratum” which means “A square figure”. Brahmagupta (an ancient Indian Mathematician )(A.D. 598-665) gave an explicit formula to solve a quadratic equation. Later Sridharacharya (A.D. 1025) derived a formula, now known as the quadratic formula, for solving a quadratic equation by the method of completing the square. An Arab mathematician Al-khwarizni(about A.D. 800) also studied quadratic equations of different types. It is believed that Babylonians were the first to solve quadratic equations. Greek mathematician Euclid developed a geometrical approach for finding lengths, which are nothing but solutions of quadratic equations.

Important Questions on Class 10 Maths Chapter 4

Check whether the following is quadratic equation: (x + 1)² = 2(x – 3).

(x + 1)² = 2(x – 3) Simplifying the given equation, we get (x + 1)² = 2(x – 3) ⇒ x² + 2x + 1 = 2x – 6 ⇒ x² + 7 = 0 or x² + 0x + 7 = 0 This is an equation of type ax² + bx + c = 0. Hence, the given equation is a quadratic equation.

Represent the following situation in the form of quadratic equation: The product of two consecutive positive integers is 306. We need to find the integers.

Let the first integer = x Therefore, the second integer = x + 1 Hence, the product = x(x + 1) According to questions, x(x + 1) = 306 ⇒ x² + x = 306 ⇒ x² + x – 306 = 0 Hence, the two consecutive integers satisfies the quadratic equation x² + x – 306 = 0.

In a class test, the sum of Shefali’s marks in Mathematics and English is 30. Had she got 2 marks more in Mathematics and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects.

Let, Shefali’s marks in Mathematics = x Therefore, Shefali’s marks in English = 30 – x If she got 2 marks more in Mathematics and 3 marks less in English, Marks in Mathematics = x + 2 Marks in English = 30 – x – 3 According to questions, Product = (x + 2)(27 – x) = 210 ⇒ 27x – x² + 54 – 2x = 210 ⇒(-x)² + 25x – 156 = 0 ⇒ x² – 25x + 156 = 0 ⇒ x² – 12x – 13x + 156 = 0 ⇒ x(x – 12) – 13(x – 12) = 0 ⇒ (x – 12)(x – 13) = 0 ⇒ (x – 12) = 0 or (x – 13) = 0 Either x = 12 or x = 13 If x = 12 then, marks in Maths = 12 and marks in English = 30 – 12 = 18 If x = 13 then, marks in Maths = 13 and marks in English = 30 – 13 = 1

The difference of squares of two numbers is 180. The square of the smaller number is 8 times the larger number. Find the two numbers.

Let the larger number = x Let the smaller number = y Therefore, y² = 8x According to question, x² – y² = 180 ⇒ x² – 8x = 180 [As y² = 8x] ⇒ x² – 8x – 180 = 0 ⇒ x² – 18x + 10x – 180 = 0 ⇒ x(x – 18) + 10(x – 18) = 0 ⇒ (x – 18)(x + 10) = 0 ⇒ (x – 18) = 0 or (x + 10) = 0 Either x = 18 or x = -10 But x ≠ -10 , as x is the larger of two numbers. So, x = 18 Therefore, the larger number = 18 Hence, the smaller number = y = √144 = 12

Sum of the areas of two squares is 468 m². If the difference of their perimeters is 24 m, find the sides of the two squares.

Let the side of larger square = x m Let the side of smaller square = y m According to question, x² + y² = 468 …(i) Difference between perimeters, 4x – 4y = 24 ⇒ x – y = 6 ⇒ x = 6 + y … (ii) Putting the value of x in equation (i), we get (y + 6)² + y² = 468 ⇒ y² + 12y + 36 + y² = 468 ⇒ (2y)² +12y – 432 = 0 ⇒ y² + 6y – 216 = 0 ⇒ y² + 18y – 12y – 216 = 0 ⇒ y(y + 18) – 12(y + 18) = 0 ⇒ (y + 18)(y – 12) = 0 ⇒ (y + 18) = 0 or (y – 12) = 0 Either y = -18 or y = 12 But, y ≠ -18 , as x is the side of square, which can’t be negative. So, y = 12 Hence, the side of smaller square = 12 m Putting the value of y in equation (ii), we get Side of larger square = x = y + 6 = 12 + 6 = 18 m

How to Revise 10th Maths Chapter 4 Quadratic Equations for Exams

Schools and institutions across the world promptly acted according to a pandemic, by moving online. Tech advancement helped institutions transition physical classrooms to virtual ones in record time. For as long as I can remember, I have liked everything about mathematics – especially teaching young schoolers, I have seen some of the brilliant minds grapple to comprehend the concepts. I have seen hard-working bright-eyed students losing to perform better than average in the classroom. In this article, you will read some of the effective practices that helped many students score 100% in math of 10th standard chapter Quadratic equations.

Step 1: NCERT Solutions for Class 10 Maths Chapter 4 helps to practice real life based Problems in Exercise 4.3.

Step 2: class 10 maths chapter 4 solutions provides the fundamental facts of quadratic equations., step 3: ncert solutions class 10 maths chapter 4 by applying the perfect formula for answers., step 4: class 10 maths chapter 4 needs regular practice session after short intervals., step 5: practice class 10 maths chapter 4 from ncert textbook for exams..

How can I get good marks in Class 10 Maths Chapter 4 Quadratic Equations?

Student should know the methods of factorization to a quadratic equation. It will help a lot during the solution of questions in 10th Maths chapter 4. Quadratic formula is the ultimate trick to find the roots of difficult or easy format of any quadratic equation. So if someone has practiced well the factorization method and quadratic formula method, he will score better then ever in chapter 4 of class 10 mathematics.

How a quadratic polynomial is different from a quadratic equation in 10th Maths Chapter 4?

A polynomial of degree two is called a quadratic polynomial. When a quadratic polynomial is equated to zero, it is called a quadratic equation. A quadratic equation of the form ax² + bx + c = 0, a > 0, where a, b, c are constants and x is a variable is called a quadratic equation in the standard format.

In Class 10 Maths Chapter 4, which exercise is considered as the most difficult to solve?

Class 10 Maths, exercise 4.1, and 4.2 are easy to solve and having less number of questions. Exercise 4.3 is tricky to find the solutions and answers also. In this exercise most of the questions are based on application of quadratic equations.

What is meant by zeros of a quadratic equation in Chapter 4 of 10th Maths?

A zero of a polynomial is that real number, which when substituted for the variable makes the value of the polynomial zero. In case of a quadratic equation, the value of the variable for which LHS and RHS of the equation become equal is called a root or solution of the quadratic equation. There are three algebraic methods for finding the solution of a quadratic equation. These are (i) Factor Method (ii) Completing the square method and (iii) Using the Quadratic Formula.

What are the main topics to study in Class 10 Maths chapter 4?

In chapter 4 (Quadratic equations) of class 10th mathematics, Students will study

• 1) Meaning of Quadratic equations
• 2) Solution of a quadratic equation by factorization.
• 3) Solution of a quadratic equation by completing the square.
• 4) Solution of a quadratic equation using quadratic formula.
• 5) Nature of roots.

How many exercises are there in chapter 4 of Class 10th Maths?

There are in all 4 exercises in class 10 mathematics chapter 4 (Quadratic equations). In first exercise (Ex 4.1), there are only 2 questions (Q1 having 8 parts and Q2 having 4 parts). In second exercise (Ex 4.2), there are in all 6 questions. In fourth exercise (Ex 4.3), there are in all 5 questions. So, there are total 13 questions in class 10 mathematics chapter 4 (Quadratic equations). In this chapter there are in all 18 examples. Examples 1, 2 are based on Ex 4.1, Examples 3, 4, 5, 6 are based on Ex 4.2, Examples 16, 17, 18 are based on Ex 4.3.

Does chapter 4 of class 10th mathematics contain optional exercise?

No, chapter 4 (Quadratic equations) of class 10th mathematics doesn’t contain any optional exercise. All the four exercises are compulsory for the exams.

How much time required to complete chapter 4 of 10th Maths?

Students need maximum 3-4 days to complete chapter 4 (Quadratic equations) of class 10th mathematics. But even after this time, revision is compulsory to retain the way to solving questions.

« Chapter 3: Pair of Linear Equations in Two Variables

Chapter 5: arithmetic progression ».

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Class 10 Mathematics Quadratic Equation Assignments

We have provided below free printable Class 10 Mathematics Quadratic Equation Assignments for Download in PDF. The Assignments have been designed based on the latest NCERT Book for Class 10 Mathematics Quadratic Equation . These Assignments for Grade 10 Mathematics Quadratic Equation cover all important topics which can come in your standard 10 tests and examinations. Free printable Assignments for CBSE Class 10 Mathematics Quadratic Equation , school and class assignments, and practice test papers have been designed by our highly experienced class 10 faculty. You can free download CBSE NCERT printable Assignments for Mathematics Quadratic Equation Class 10 with solutions and answers. All Assignments and test sheets have been prepared by expert teachers as per the latest Syllabus in Mathematics Quadratic Equation Class 10. Students can click on the links below and download all Pdf Assignments for Mathematics Quadratic Equation class 10 for free. All latest Kendriya Vidyalaya Class 10 Mathematics Quadratic Equation Assignments with Answers and test papers are given below.

Mathematics Quadratic Equation Class 10 Assignments Pdf Download

We have provided below the biggest collection of free CBSE NCERT KVS Assignments for Class 10 Mathematics Quadratic Equation . Students and teachers can download and save all free Mathematics Quadratic Equation assignments in Pdf for grade 10th. Our expert faculty have covered Class 10 important questions and answers for Mathematics Quadratic Equation as per the latest syllabus for the current academic year. All test papers and question banks for Class 10 Mathematics Quadratic Equation and CBSE Assignments for Mathematics Quadratic Equation Class 10 will be really helpful for standard 10th students to prepare for the class tests and school examinations. Class 10th students can easily free download in Pdf all printable practice worksheets given below.

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Assignments Class 10 Mathematics Quadratic Equation Pdf Download

Students can refer to Assignments for Class 10 Mathematics Quadratic Equation available for download in Pdf. We have given below links to subject-wise free printable Assignments for Mathematics Quadratic Equation Class 10 which you can download easily. All assignments have a collection of questions and answers designed for all topics given in your latest NCERT Books for Class 10 Mathematics Quadratic Equation for the current academic session. All Assignments for Mathematics Quadratic Equation Grade 10 have been designed by expert faculty members and have been designed based on the type of questions asked in standard 10 class tests and exams. All Free printable Assignments for NCERT CBSE Class 10, practice worksheets, and question banks have been designed to help you understand all concepts properly. Practicing questions given in CBSE NCERT printable assignments for Class 10 with solutions and answers will help you to further improve your understanding. Our faculty have used the latest syllabus for Class 10. You can click on the links below to download all Pdf assignments for class 10 for free. You can get the best collection of Kendriya Vidyalaya Class 10 Mathematics Quadratic Equation assignments and questions workbooks below.

Class 10 Mathematics Quadratic Equation Assignments Pdf Download

CBSE NCERT KVS Assignments for Mathematics Quadratic Equation Class 10 have been provided below covering all chapters given in your CBSE NCERT books. We have provided below a good collection of assignments in Pdf for Mathematics Quadratic Equation standard 10th covering Class 10 questions and answers for Mathematics Quadratic Equation. These practice test papers and workbooks with question banks for Class 10 Mathematics Quadratic Equation Pdf Download and free CBSE Assignments for Class 10 are really beneficial for you and will support in preparing for class tests and exams. Standard 10th students can download in Pdf by clicking on the links below.

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• We have provided step by step solutions for all questions in the Class 10 assignments so that you can understand the solutions in detail.
• We have provided single click download links to all chapterwise worksheets and assignments in Pdf.
• Class 10 practice question banks will support to enhance subject knowledge and therefore help to get better marks in exam

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Assignments For Class 10 Mathematics Quadratic Equation

Assignments for Class 10 Mathematics Quadratic Equation have been developed for Standard 10 students based on the latest syllabus and textbooks applicable in CBSE, NCERT and KVS schools. Parents and students can download the full collection of class assignments for class 10 Mathematics Quadratic Equation from our website as we have provided all topic wise assignments free in PDF format which can be downloaded easily. Students are recommended to do these assignments daily by taking printouts and going through the questions and answers for Grade 10 Mathematics Quadratic Equation. You should try to do these test assignments on a daily basis so that you are able to understand the concepts and details of each chapter in your Mathematics Quadratic Equation book and get good marks in class 10 exams.

Short Answer type Questions :

Question. Solve the following quadratic equation for x: 4x 2  + 4bx – (a 2  – b 2 ) = 0.  Solution.  4x 2  + 4bx – (a 2  – b 2 ) = 0

Question. Solve the following quadratic equation for x: x 2  – 2ax – (4b 2  – a 2 ) = 0. Solution.  x 2  – 2ax – (4b 2  – a 2 ) = 0

Question. 2√3x 2  – 5x + √3 = 0 Solution.  √3 /2 , 1/√3

Question. 3x 2  + 2√5x – 5 = 0 Solution.  √5/3 , – √5

Question. 1/ x + 4 – 1/x – 7 = 11/30 , x ≠ – 4, 7 Solution.  2, 1

Question. Find the roots of quadratic equation: x 2  − 3√5x +10 = 0    Solution.   

Question. Write the discriminant of the quadratic equation (x + 5) 2  = 2(5x –3). Solution.  (x + 5) 2  = 2(5x – 3) ⇒ x 2  + 25 + 10x = 10x – 6 ⇒ x 2  + 31 = 0 ⇒ x 2  + 0x + 31 = 0 ∴ D = (0) 2  – 4 × 1 × 31 = 0 – 124 = –124

Question. Solve the following quadratic equation for x: 9x 2  – 6b 2 x – (a4 – b4) = 0.    Solution.  9x 2  – 6b 2 x – (a 4  – b 4 ) = 0

Question. Find the roots of the equation ax 2  + a = a 2 x + x.  Solution.  a , 1/a

Question. Find the roots of quadratic equation: 5√5x 2  + 30x + 8√5  Solution.

Question. Solve for x: x – 3/x – 4  + x – 5/ x – 6 = 10/6 ; x ≠ 4,6 Solution.  2 ± √10

Question. Solve for x: x – 2 /x – 3 + x – 4/x – 5 = 10/3 ; x ≠ 3 , 5 Solution.  7/2 , 6

Question. Solve for x: 4x 2  – 4ax + (a 2  – b 2 ) = 0.  Solution.

Question. Two water taps together can fill a tank in 9 hours 36 minutes. The tap of large diameter takes 8 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank. Solution.  Let x be the time taken by larger diameter tap.          ∴  x + 8 be the time taken by smaller diameter tap.

x = 16 or x = – 4.8 (Rejected) Hence, time taken by larger and smaller taps are 16 hrs and 24 hrs respectively.

Question. Solve for x: 3(3x – 1/2x + 3) – 2 (2x + 3/3x – 1) = 5 ; x ≠ 1/3 , – 3/2 Solution.  0, –7

Question. The difference of squares of two numbers is 88. If the larger number is 5 less than twice the smaller number, then find the two numbers.  Solution.  9 and 13

Question. Solve the following quadratic equation for x: 4x 2  – 4a2x + (a 4 – b 4 ) = 0.        Solution.  4x 2  – 4a 2 x + (a 4  – b 4 ) = 0

Question. Find the discriminant of the quadratic equation: 4x 2  – 2/3x – 1/16 = 0  Solution.  3328

Question. Sum of the areas of two squares is 544 m 2 . If the difference of their perimeters is 32 m, find the sides of the two squares. Solution.  Let the sides of two squares in metres be x and y respectively (where x > y).      Given: Sum of areas of two squares = 544 m 2 ⇒ x 2  + y 2  = 544 …(i)  Also, difference of their perimeters = 32 m ⇒ 4x – 4y = 32 ⇒ x – y = 8 ⇒ y = x – 8 …(ii) (1) Substituting the value of y for equation (ii) in equation (i), we get x 2  + (x – 8) 2  = 544 ⇒ x 2  + x 2  – 16x + 64 – 544 = 0 ⇒ 2x 2  – 16x – 480 = 0 ⇒ x 2  – 8x – 240 = 0

Question. A rectangular field is 20 m long and 14 m wide. There is a path of equal width all around it, having an area of 111 sq m. Find the width of the path. Solution.  1.5 m Question. At ‘t’ minutes past 2 pm, the time needed by the minute hand of a clock to show 3 pm was found to be 3 minutes less than t2/4 minutes. Find t. Solution.  14 minutes

Question. Find a natural number whose square diminished by 84 is equal to thrice of 8 more than the given number. Solution.  

Question. Solve for x: √3x 2  + 14x – 5√3 = 0 Solution.  

Question. Find the roots of the equation x 2  + 7x + 10 = 0. Solution.  x 2  + 7x + 10 = 0 x 2  + 5x + 2x + 10 = 0 (½) (x + 5)(x + 2) = 0 x = –5, x = –2

Question. Solve for x: 4√3x 2  + 5x – 2√3  = 0 Solution.  Consider   

Question. Solve for x: x 2  x – (√2 + 1) x + √2 = 0 Solution.  Consider:   

Question. Solve for x: 1/ x – 3  – 1/x + 5 = 1/6 , x ≠ 3, – 5. Solution.   

Question. Solve the quadratic equation 2x 2 + ax – a 2 = 0 for x using quadratic formula.      Solution. 2x 2 + ax – a 2 = 0 Here, a = 2, b = a and c = –a 2 . Using the formula,

Question. Solve the following quadratic equation:      9x 2 – 9 (a + b) x + [2a 2 + 5ab + 2b 2 ] = 0  Solution. We have 9x 2 – 9 (a + b) x + [2a 2 + 5ab + 2b 2 ] = 0 Here, A = 9, B = –9 (a + b) and C = [2a 2 + 5ab + 2b 2 ] So, discriminant, D = B 2 – 4 AC = { –9 (a + b)} 2 – 4 × 9 (2a 2 + 5 ab + 2b 2 ) = 9 2 (a + b) 2 – 4 × 9 (2a 2 + 5ab + 2b 2 ) = 9 {9 (a + b) 2 – 4 (2a 2 + 5 ab + 2b2 )} = 9 {9a 2 + 9b 2 + 18ab – 8a 2 – 20ab – 8b 2 } = 9 (a 2 + b 2 – 2ab) = 9 (a – b)

Question. Find the solution of the quadratic equations by quadratic formula.   

Question. Using the quadratic formula, solve the equation      a 2 b 2 x 2 – (4b 4 – 3a 4 ) x – 12a 2 b 2 = 0  Solution. Comparing given equation with Ax 2 + Bx + C = 0, we get A = a 2 b 2 , B = –(4b 4 – 3a 4 ) and C = –12a 2 b 2 . ∴ B 2 – 4AC = (4b 4 – 3a 4 ) 2 – 4 × a 2 b 2 × (–12a 2 b 2 ) = 16b 8 + 9a 8 – 24a 4 b 4 + 48a 4 b 4 = 16b 8 + 9a8 + 24a 4 b 4 = (4b4 + 3a4) 2

Question. Two water taps together can fill a tank in 6 hours. The tap of larger diameter takes 9 hours less than the smaller one to fill the tank separately. Find the time in which each tap can separately fill the tank. Solution.  9 hrs.

Question. Two pipes running together can fill a tank in 11 ,1/9 minutes. If one pipe takes 5 minutes more than the other to fill the tank separately, find the time in which each pipe would fill the tank separately . Solution.  20 minutes, 25 minutes.

Question. The difference of two natural numbers is 3 and the difference of their reciprocals is 3/28 . Find the numbers.  Solution.  4, 7

Question. √3x 2  + 10x + 7√3 = 0 Solution.  -√3 , -7/√3 

Question. Solve for x: x + 1 /x – 1 + x – 2/x + 2 = 4 – 2x + 3/x – 2 ; x ≠ 1 , -2 , 2 . Solution.  

Question. (x – 3)(2x + 3) = 0 Solution.  3, -3/2

Question. A pole has to be erected at a point on the boundary of a circular park of diameter 17 m in such a way that the differences of its distances from two diametrically opposite fixed gates A and B on the boundary is 7 metres. Find the distances from the two gates where the pole is to be erected.  Solution.  Let P be the position of the pole. ∠APB = 90° (angle in a semicircle)   

Question. 3x 2  – 2ax – a 2  = 0 Solution.  a , -a/3

Question. At present Asha’s age (in years) is 2 more than the square of her daughter Nisha’s age. When Nisha grows to her mother’s present age, Asha’s age would be one year less than 10 times the present age of Nisha. Find the present ages of both Asha and Nisha.  Solution.  Nisha’s age = 5 years, Asha’s age = 27 years

Question. The difference of two natural numbers is 5 and the difference of their reciprocals is 5/14 . Find the numbers. Solution.  7, 2

Question. Solve the equation for x: 3x – 4 / 7 + 7/3x – 4 = 5/2 , x ≠ 4/3 . Solution.   

Question. Find two consecutive positive integers sum of whose squares is 365.  Solution.  Let two consecutive positive integers be x and x + 1 ∴ x 2  + (x + 1) 2  = 365  ⇒ x 2  + x − 182 = 0  (x + 14)(x − 13) = 0  ∴ x = 13  Hence, two consecutive positive integers are 13 and 14.

Question. A train travels at a certain average speed for a distance of 63 km and then travels at a distance of 72 km at an average speed of 6 km/hr more than its original speed. If it takes 3 hours to complete total journey, what is the original average speed? Solution.  Let the original average speed of train be x km/hr.   

Question. 3a 2 x 2  + 8abx + 4b 2  = 0 Solution.  – 2b/a , – 2b/3a

Question. Some students planned a picnic. The total budget for food was ₹ 2,000. But 5 students failed to attend the picnic and thus the cost of food for each member increased by ₹ 20. How many students attended the picnic and how much did each student pay for the food?  Solution.  Case I. Let number of students = x and cost of food for each member = ₹ y Then x × y = 2,000 …(i)(1) Case II. New number of students = x – 5 New cost of food for each member = ₹ (y + 20) Then (x – 5)(y + 20) = 2,000 fi xy + 20x – 5y – 100 = 2,000 …(ii)(1) Solving (i) and (ii), we get ∴ x = –20, 25 (1) x = –20 is rejected because number of students can’t be negative. So, x = 25 ∴ y = 80  Number of students = 25 Cost of food for each student = ₹ 80.

Question. Solve the equation for x: 1/x + 1  + 2/x+2 = 5/x +4 , x ≠ -1 , -2 , -4 . Solution.  Given that:   

Question. A two-digit number is such that the product of its digits is 14. When 45 is added to the number, the digits interchange their places. Find the number.  Solution.  27

Question. A motorboat whose speed in still water is 18 km/h, takes 1 hour more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream. Solution.  6 km/h.

Question. Solve for x: √2x + 9 + x = 13. Solution.   

Question. In a flight of 600 km, an aircraft was slowed down due to bad weather. The average speed of the trip was reduced by 200 km/hr and the time of flight increased by 30 minutes. Find the duration of flight. Solution.  1 hr 

Question. Solve for x: √6x + 7 − (2x − 7) = 0. Solution.   

Question. A rectangular park is to be designed whose breadth is 3 m less than its length. Its area is to be 4 square metres more than the area of a park that has already been made in the shape of an isosceles triangle with its base as the breadth of the rectangular park and of altitude 12 m. Find the length and breadth of the park.  Solution.  Let ABCD is a rectangular park and CDE is a triangular park (isosceles triangle).     

Question. Solve for x: √3x 2  – 2√2x – 2√3 = 0 Solution.     

Question. Solve the following equation: 1//x – 1/x – 2 = 3 , x ≠ 0, 2 . Solution.   

Quadratic Equation :  a polynomial of second degree is called quadratic equation. ax 2  +bx +c,a≠0

Methods for solving quadratic equation

• Factorization method
• Completing the square
• Quadratic formula
• Foots of quadratic equation ax 2  +bx +c=0 are

Nature of roots :

• Two distinct real roots, if b 2 – 4ac >0,
• Two equal real roots, if b 2 -4ac=0 and each root is b/2a
• No real roots, ifif b 2 – 4ac < 0.if b 2 – 4ac is called discriminant of the quadratic equation and denoted by D.
• A quadratic polynomial whose zeroes are α,β is given by p(x) = x 2  =(α + β)x + αβ =x 2 -( sum of the zeroes)x+product of the zeroes

Cubic polynomial whose zeroes are α,β,γ is given by p(x) = x 3  -(α + β +γ)x 2  +( αβ + βγ + αγ)x – α β γ = x 3  -( sum of the zeroes)x 2  +( sum of the products of the zeroes taken two at a time)-product of the zeroes

Question. Find the value/s of k for which the quadratic equation 3x 2  + kx + 3 = 0 has real and equal roots. Solution.  Given, quadratic equation is: 3x 2  + kx + 3 = 0 For real and equal roots b 2  – 4ac = 0 Here, a = 3, b = k and c = 3 ⇒ b 2  – 4ac = (k) 2  – 4 × 3 × 3 = 0 ⇒ k 2  = 36 ⇒ k = ± 6 Hence, the value of k is –6.

Question. Solve for x :

Question. The sum of the squares of two consecutive odd numbers is 394. Find the numbers. Solution.  Let the two consecutive odd numbers be x and x + 2. ∴ x 2  + (x + 2) 2  = 394 ⇒ x 2  + x 2  + 4 + 4x = 394 ⇒ 2x 2  + 4x + 4 = 394 ⇒ 2x 2  + 4x – 390 = 0 ⇒ x 2  + 2x – 195 = 0 ⇒ x 2  + 15x – 13x – 195 = 0 ⇒ x (x + 15) – 13 (x + 15) = 0 ⇒ (x – 13) (x + 15) = 0 Either x – 13 = 0 or x + 15 = 0 ⇒ x = 13 or x = – 15 (neglected) When first number x = 13, then second number x + 2 = 13 + 2 = 15.

Question. Two pipes running together can fill a cistern in 3,1/13 hours. If one pipe takes 3 hours more than the other to fill it, find the time in which each pipe would fill the cistern. Solution.  Let time taken by faster pipe to fill the cistern be x hrs. Therefore, time taken by slower pipe to fill the cistern = (x + 3) hrs Since the faster pipe takes x minutes to fill the cistern. ∴ Portion of the cistern filled by the faster pipe in one hour = 1/x  Portion of the cistern filled by the slower pipe in one hour = 1/x + 3 Portion of the cistern filled by the two pipes together in one hour = 1/40/30 = 13/40

Question. The sum of two numbers is 15 and the sum of their reciprocals is 3/10 . Find the numbers. Solution.  Let the numbers be x and 15 – x. According to given condition,

Question. Solve the following quadratic equations by the factorisation method. (a) 7x 2  = 8 – 10x (b) x (x + 9) = 52 (c) 3 (x 2  – 4) = 5x (d) x (x + 1) + (x + 2) (x + 3) = 42 Solution.  (a) 7×2 = 8 – 10x ⇒ 7x 2  + 10x – 8 = 0 ⇒ 7x 2  + 14x – 4x – 8 = 0 ⇒ 7x(x + 2) – 4 (x + 2) = 0 ⇒ (7x – 4) (x + 2) = 0

Question. Solve for x:

Question. If Zeba was younger by 5 years than what she really is, then the square of her age (in years) would have been 11 more than five times her actual age. What is her age now? Solution.  Let the present age of Zeba be x years. Age before 5 years = (x – 5) years According to given condition, (x – 5) 2  = 5x + 11 ⇒ x 2  + 25 – 10x = 5x + 11 ⇒ x 2  – 10x – 5x + 25 – 11 = 0 ⇒ x 2  – 15x + 14 = 0 ⇒ x 2  – 14x – x + 14 = 0 ⇒ x (x – 14) – 1 (x – 14) = 0 ⇒ (x – 1) (x – 14) = 0 ⇒ x – 1 = 0 or x – 14 = 0 ⇒ x = 1 or x = 14 But present age cannot be 1 year. ∴ Present age of Zeba is 14 years. 

Question. Solve the equation

Solution. Here,

Question. The difference of two natural numbers is 5 and the difference of their reciprocals is 1/10 . Find the numbers. Solution.  Let the two natural numbers be x and y such that x > y. According to the question, Difference of numbers, x – y = 5 ⇒ x = 5 + y …(i) Difference of their reciprocals,

Question. Speed of a boat in still water is 15 km/h. It goes 30 km upstream and returns back at the same point in 4 hours 30 minutes. Find the speed of the stream. Solution.  Let the speed of stream be x km/hr. ∴ Speed of boat in upstream = (15 – x) km/hr. Speed of boat in downstream = (15 + x) km/hr.

Question. A train travelling at a uniform speed for 360 km would have taken 48 minutes less to travel the same distance if its speed were 5 km/hour more. Find the original speed of the train. Solution.  Let original speed of the train be x km/hr. Time taken at original speed = 360/x hours

Question. A plane left 30 minutes late than its scheduled time and in order to reach the destination 1500 km away in time, it had to increase its speed by 100 km/h from the usual speed. Find its usual speed. Solution.  Let the usual speed of the plane be x km/hr.

Question. In the centre of a rectangular lawn of dimensions 50 m × 40 m, a rectangular pond has to be constructed so that the area of the grass surrounding the pond would be 1184 m 2 . Find the length and breadth of the pond.    Solution.  Let ABCD be a rectangular lawn and EFGH be rectangular pond. Let x m be the width of grass area, same around the pond. Now, length of lawn = 50 m, width of lawn = 40 m ∴ Length of pond = (50 – 2x) m, width of pond = (40 – 2x) m Since area of grass surrounds the pond = 1184 m 2 ⇒ Area of lawn – Area of pond = 1184 m 2 ⇒ 50 × 40 – (50 – 2x) (40 – 2x) = 1184

⇒ 2000 – (2000 – 80x – 100x + 4x 2 ) = 1184 ⇒ 4x 2  – 180 x + 1184 = 0 or x2 – 45x + 296 = 0 fi x 2  – 37x – 8x + 296 = 0 ⇒ (x – 37) (x – 8) = 0 fi x = 37, 8 Since x = 37 is not possible, as otherwise the length of pond will be negative. Hence, x = 8 is the required solution. ∴ Length of pond = 50 – 2 × 8 = 34 m and breadth of pond = 40 – 2 × 8 = 24 m

Question. For what values of k does the quadratic equation 4x 2  – 12x – k = 0 have no real roots ? Solution.  Given equation is: 4x 2  – 12x – k = 0 For equation to have no real roots, D < 0 or b 2  – 4ac < 0 Here, a = 4, b = – 12, c = – k (– 12) 2  – 4 × 4 × (– k) < 0 144 + 16k < 0 16k < –144 k < –9 Hence, the value of k should be less than – 9.

Question. Find the nature of the roots of the quadratic equation 2x 2  – 4x + 3 = 0.  Solution.  Given: quadratic equation : 2x 2  – 4x + 3 = 0 Here, a = 2, b = – 4, c = 3 Discriminant, D = b 2  – 4ac = (– 4) 2  – 4 × 2 × 3 = 16 – 24 = – 8 < 0 ∴ D < 0 Hence, the given equation does not have real roots.

PRACTICE EXERCISE

Question. 25x 2  – 10 x – 8 = 0 Solution.  4/5 , -2 /5

Question. x 2  – 3x – 28 = 0 Solution.  7, –4

Question. 3x 2  + 12ax – a 2  = 0, where a is real. Solution.  – a , a/3

Question. 5x 2  + 16x = – 12 Solution.  – 2 , 6/5

Question. 6 √2x 2  + 5x – 3 √2 = 0 Solution.  √2/3 , – 3/2√2

Question. 6a 2 x 2  – 7abx – 3b 2  = 0 Solution.  – b /3a , 3b/2a

Question. x 2  – (√3 +1)x + √3 = 0 Solution.  1, √3

Question. 4x 2  + 3x + 9 = 0 Solution.  No real roots

Question. 4x 2  + 4 √3 x + 3 = 0 Solution.  – √3/2 , – √3/2

Question. – 8x 2  +10x = 3 Solution.  3/4 , 1/2

Question. 15x 2  – 28 = x Solution.  7/5 , -4/3

Question. x 2  – 2ax + a 2  – b 2  = 0 Solution.  a – b, a + b

Question. 4x 2  – 4a 2 x + (a 4  – b 4  ) = 0 Solution.  a 2  + b 2  /2 , a 2  – b 2  /2

Question. If x = 5 and x = 5/4 are the roots of the equation ax2 – 15x + b = 0. Find the value of a and b. Solution.  a = 4, b = – 25

Question. Determine whether x = 1/2 and x = 3/2 are solutions of the equation 2x 2  -5x +3 = 0 or not. Solution.  x =1/2 is not a solution, but x = 3/2 is a solution.

Question. 3x 2  – 5x + 2 = 0 Solution.  1, 2/3

Question. a 2 b 2 x 2  + b2x – a 2 x +1 = 0 Solution.  -1/a 2  , 1/b 2

Question. 1/ a + b + x = 1/a + 1/b + 1/x , a + b ≠ 0 Solution.  –a, –b

Question. x + 3 / x – 2 – 1 – x/x = 17/4   Solution.  4 , -2/9

Question. x – 1 /2x – 1 + 2x + 1 /x – 1 = 5/2 ; x ≠ – 1/2,1 Solution.  –1

Question. x – 1 /x – 2 + x – 3 / x – 4 =  3 , 1/3 ; x ≠ 2,4 Solution.  5 , 5/2

Question. x 2  – 8x = 0 Solution.  0, 8

Question. 2x 2  + x – 4 = 0 Solution.  – + √33 / 2 , -1- √33 / 4

Question. 9x 2  – 15x + 6 = 0 Solution.  1,2/3

Question. x 2  – 6 √2x + 10 = 0 Solution.  √2,5 √2

Question. x 2  – (√2 +1)x + √2 = 0 Solution.  √2, 1

Question. a 2 x 2  – 3abx + 2b 2  = 0 Solution.  2b/a , b/a

Question. 3x 2  – 5x +2 = 0 Solution.  1

Question. Which of the following are quadratic equations ? (a) x 2  + 7x – 12 = 0 (b) 9x 2  – 12 x = 0  (c) x – 4/x = x 2 (d) 7x 2  = 0 (e) x 2  + 1/x 2  = 4 (f) x(x + 7) = x 2  – 4x + 2 Solution.  (a), (b), (d)

Question. In each of the following, determine whether the given values are the solutions of the given quadratic equations or not : (a) x 2  – 5x + 6 = 0 ; x = 2, x = 3 (b) 7×2 + 4x – 20 = 0; x = – 2 , x = – 10/7 (c) x + 1/x = 25/12 ; x = 2/3 , x = 3/4  (d) x 2  – 2 √3 – 9 = 0 ; x = – 3, x = 3 √3 (e) √3 x 2  + 11 x + 6√3 = 0 ; x = – 2/√3 , x = – 3√3 Solution.  (a) x = 2 and x = 3 are solutions. (b) x = – 2 is a solution, but x = – 10/7 is not a solution. (c) x = 2/3 is not a solution, x = 3/4 is a solution. (d) x = – √3 and x = 3 √3 are solutions. (e) x = -2/√3 and x = =3 √3 are solutoins.

Question. –4x 2  = 6 – 3x Solution.  –87

Question. x 2  – 9x + 20 = 0 Solution.  4, 5

Question. x 2  – 6x + 9 = 0 Solution.  3, 3

Question. x 2  – 2xb + b 2  = 0 Solution.  b, b

Question. x 2  – 4x + 4 = (3/2) 2 Solution.  7/2 , 1/2

Question. 3x 2  + 2 √5x – 5 = 0 Solution.  -√5 , √5/3

Question. √5x 2  + 9x -14 √5 = 0 Solution.  √5 , -14/√5

Question. 2x 2  – (q + 2 p)x + pq = 0 Solution.  p , q/2

Question. 3y 2  + (6 + 4a) y + 8a = 0 Solution.  – 2, -4a/3

Question. a/x – b + b/x – a = 2 ; x ≠ a , b Solution.  a + b , a + b /2

Question. 4x 2  – 2 (a 2  + b 2 ) x + a 2 b 2  = 0 Solution.  a 2  /2 , b 2 /2

Question. (k + 4)x 2  + (k + 1)x + 1 = 0 Solution.  k = 0 or 3

Question. 5x 2  – 6x + p = 0 Solution.  p ≤ 9/5

Question. abx 2  – (a + b)2 (x – 1) = 0 Solution.  a + b /2 , a + b /a

Question. 9x 2  – 9(a + b)x + (2a 2  + 5ab + 2b 2 ) = 0 Solution.  2a + b /3 , a + 2b/3

Question. abx 2  + (b 2  – ac)x – bc = 0 Solution.  – b/a , c/b

Question. x 2  + x(a + 2) + (a +1) = 0 Solution.  –1, –(1 + a)

Question. 4/x – 3 = 5/2x + 3 ; x ≠ 0 , – 3/2 Solution.  –2, 1

Question. p 2  x 2  + ( p2 – q 2  )x – q2 = 0 Solution.  – 1, q 2/ p 2

Question. 1/ x + 1 + 2/x + 2 = 4/x + 4 ; x ≠ 1 ,  2, 4 Solution.  2 ± 2 √3

Question. 6x 2  + (12 – 8a)x -16a – 0 Solution.  – 2 , 4a/3

Question. 2x – 3 /x – 2 + 2x – 7/x – 4 = 16/3 ; x ≠ 2,4 Solution.  5, 5/2

Question. x -a/x – b  + x – b /x -a = a/b + b/a ; x ≠ a,b Solution.  0, a + b

Question. In each of the following, find the value of p for which the given value is a solution of the given equation: (a) (2 p +1)x 2  + 2x – 3 = 0; x = 2 (b) x 2  + 2ax – 2 p = 0; x = -a (c) x 2  – x(a + b) + p = 0; x = b (d) px 2  – √2x – 4 = 0; x = √2 Solution.  (a) p = -5/8 (b) p = – a 2 /2 (c) p = ab (d) p = 3

Question. 5x 2  + 4x – 1 = 0 Solution.  36

Question. x 2  + 2ax + 3b = 0 Solution.  4a 2  – 12b

Question. √2x 2  – 3x – 3 √2 = 0 Solution.  33

Question. 9x 2  – 6x + 1 = 0 Solution.  Real and equal

Question. x 2  + x – 12 = 0 Solution.  Real and unequal

Question. 4√3x 2  + 5x + 2 √3 = 0 Solution.  no real roots

Question. 4x 2  – 4x + 1 = 0 Solution.  Real and equal

Question. (x – 2a) (x – 2b) = 4ab Solution.  Real and unequal

Question. 9x 2  – 5 px + 1 = 0 Solution.  p ≥ 6/5 or p ≤ – 6/5

Question. x 2  + 4x + p = 0 Solution.  p ≤ 4

Question. 2x 2  + px + 2 = 0 Solution.  p ≤ – 4 or p ≥ 4

Question. 4x 2  – 3px + 1 = 0 Solution.  p ≤ – 4/3 or p ≥ 4/3

Question. For what value of k, (4 – k) x 2 + (2k + 4)x + (8k + 1) = 0, is a perfect square. Solution.  k = 0, 3

Question. 2x 2  – 10x + k = 0 Solution.  k = 25/2

Question. kx 2  – 2 √5 x + 4 = 0 Solution.  k = 5/2

Question. (k + 4)x 2  + (k + 1)x + 1 = 0 Solution.  k = – 3 or 5

Question. k  2  x 2  – 2(2k -1)x + 4 = 0 Solution.  k = 1/4

Question. Is 0.2 a root of the equation x 2  – 0.4 = 0?  Solution.  No, because 0.2 does not satisfy the quadratic equation i.e. x 2  – 0.4 = (0.2) 2  – 0.4 = 0.04 – 0.4 = -0.36 ≠ 0

Question. Find the Value of K if the equation x 2-  2(k+1)x + k 2  = 0 has equal roots. Solution.  ± 1

Question. Find the roots of x 2 – 3x -10 = 0 Solution.  -2 and 5

Question. A two digit number is such that the product of its digit is 35.When18 is added to the number, the digits interchange the places. Find the number. Solution.  57

Question. Solve 3x 2 -23x-110 = 0 Solution.  -10/3, 11

Question. Solve the following equation for ‘x’, 9x 2 -9(a+b)x+2a 2 +5ab+2b 2  = 0 Solution.  a + b/3 , 2(a + b)/3

Question. The product of two consecutive odd numbers is 483.Find the numbers. Solution.  21,23

Question. If the x = 2 and x = 3 are roots of the equation 3x 2  + 2 kx − 2m = 0, find the value of k and m. Solution.  m = 9 and k=-15/2

Question. Solve the equation: x/x + 1  + x + 1/x  = 34/15x ≠ 0; x ≠ -1 . Solution.  3/2 or -5

Question. Solve the equation 2x 2  – 5 x + 3= 0 by the method of completing the square. Solution.  3/2 or 1

Question. Using quadratic formula, solve the equation: p 2 x 2  + (p 2 – q 2 )x – q 2  = 0 Solution.  q 2 /p 2  or -1

Question. 300 apples are distributed equally among a certain number of student’s .Had there been 10 more Students, each would have received one apple less. Find the number of students. Solution.  50

Question. Find the roots of Quadratic equation 16x 2 –24x –1= 0 by using the quadratic formula. Solution.  3 ± √10/4

Question. Find the Discriminant of the Quadratic equation 2x 2 -4x+3=0and hence find the nature of its roots. Solution.  -8, and roots are not real.

Question. In a class test ,the sum of Shefali‘s marks in math’s and English is 30.Had she got 2  marks more in math’s and 3 marks less in English, the product of their marks would have been 210. Find her marks in the two subjects. Solution.  (Marks in math’s=12, marks in English=18) or (marks in math’s=13, marks in English=17)

Question. Represent the situation in the form of Quadratic equation: The Product of Rahman’s age (in years) 5 years ago with his age 9 years later is 15. Solution.  x 2 +4x-60 =0

Question. A person on tour hasRs.360 for his daily expenses. If he exceeds his tour Programme by four days, he must cut down his daily expenses, by Rs 3 per day. Find the number of days of his tour Programme. Solution.  20 days.

Question. Find the value of ‘p’ so that the equation 3x 2  – 5x – 2p = 0has equal roots. Also find the roots. Solution.  25/24

Question. The sum of two numbers is 15. If the sum of their reciprocals is 3/10 . Find the two numbers. Solution.  (10,5)or(5,10)

Question. Find the quadratic equation whose roots are 2 + √3and 2  − √3. Solution.  x 2 -4x +1=0

Question. For what value of k, x = a is a solution of equation x 2  -(a+b)x+k=0? Solution.  K=ab

Question. Find the values of K for which the .equation 9x 2  + 2kx +1 = 0 have real roots. Solution.  -3≤ k ≤ 3

Question. Divide 29 intotwopartssothatthesumofsquaresofthepartsis 425. Solution.  (13,16) or (16,13)

Question. If D > 0,then write the roots of a quadratic equation ax 2 +bx+c = 0 Solution.  -b ± √b 2  – 4ac/2a

Question. Find the Discriminant of x 2 + 5x+ 5 = 0. Solution.  5

Question. Find the sum of roots of a quadratic equation x 2  + 4x- 32 = 0 Solution.  -4

Question. Find the product of the roots of the quadratic equation 2x 2  + 7x – 4 = 0 Solution.  – 2

Question. If the equation (1 +m 2 )x 2  – 2mcx – c 2  – a 2  = 0 has equal roots. Show that c 2  = a 2 (1 + m 2 ). Solution.  Rs 58 per litre

Question. If the price of petrol is increased by Rs.2 per litre,a person had to buy1 litre less petrol for Rs.1740. Find the Original price of the petrol at that time. a) Why do you think the price of petrol is increasing day by day? b) What should we do to save petrol? Solution.  Yes,l=20mandb=20m.

Question. Solve for x: 9a 2  – 6ax – (a 2  – b 2 ) = 0 Solution.  (a+b)/3 and (a- b)/3

Short Answer Type Questions 

Question. If x = − 1/2 is a solution of the quadratic equation 3x 2 + 2kx – 3 = 0, find the value of k. 

Question. Find the value of k for which x = 3 is a solution of the equation kx 2 x + √3 − 4 = 0.  Solution.  

Question.. A cottage industry produces a certain number of toys in a day. The cost of production of each toy (in rupees) was found to be 55 minus the number of toys produced in a day. On a particular day, the total cost of production was ₹ 750. We would like to find out the number of toys produced on that day. Solution. x 2 – 55x + 750 = 0

Question.. If one root of the quadratic equation 3x 2 + px + 4 = 0 is 2/3 , then find the value of p and the other root of the equation. 8. 3x 2 + px + 4 = 0 3(2/3) 2  + P(2/3) + 4 = 0 4/3 + 2P/3 + 4 = 0 (½) p = –8 (½) 3x 2 – 8x + 4 = 0 3x 2 – 6x – 2x + 4 = 0 (½) x = 2/3 or x = 2 (½) Hence, x = 2 (½)

Question.. If x = 2/3 and x = –3 are roots of the quadratic equations ax2 + 7x + b = 0, find the values of a and b.  Solution.

Question.. John and Jivanti together have 45 marbles. Both of them lost 5 marbles each, and the product of the number of marbles they have now is 124. We would like to find out how many marbles they had to start with. Solution. x 2 – 45x + 324 = 0

Question. Solve the given quadratic equation for x: 9x 2 – 9(a + b)x + (2a 2  + 5ab + 2b 2 ) = 0 Solution.  9x 2  – 9(a + b)x + (2a 2  + 5ab + 2b 2 ) = 0 On comparing the given equation with ax 2  + bx + c = 0, we get : a = 9, b = – 9(a + b), c = 2a 2 + 5ab + 2b 2 Discriminant, D = b 2  – 4ac = [– 9(a + b)] 2  – 4 × 9(2a 2  + 5ab + 2b 2 ) = 81(a + b) 2  – 36(2a 2  + 5ab + 2b 2 ) = 81(a 2  + b 2  + 2ab) – 72a 2 – 180ab – 72b 2 = 81a 2  + 81b 2  + 162ab – 72a2 – 180ab – 72b 2 = 9a 2  + 9b 2  – 18ab = 9(a 2  + b 2  – 2ab) = 9(a – b) 2  = [3(a – b)] 2 Then, roots : x = – b ± √D /2a

Question. For what values of k, the roots of the equation x 2  + 4x + k = 0 are real? Solution.  Since, the roots of the equation x 2  + 4x + k = 0 are real, i.e. D ≥ 0 b 2  – 4ac ≥ 0 Here, a = 1, b = 4, c = k ⇒ (4) 2  – 4 × 1 × k ≥ 0 ⇒ 16 – 4k ≥ 0 ⇒ k ≤ 4 Hence, the value of ‘k’ is less than or equal to 4.

Question. If x = 2 and m = 3, the equation is 3x 2  – 2kx + 2m = 0, find k.  Solution.  3x 2  – 2kx + 2m = 0 x = 2 and m = 3 [Given] So, 3(2) 2  – 2k(2) + 2(3) = 0 ⇒ 12 – 4k + 6 = 0 ⇒ – 4k + 18 = 0 ⇒ k = 9/ 2.

Question. If one root of the quadratic equation 6x 2  – x – k = 0 is 2/ 3 , then find the value of ‘k’. Solution.  Given: quadratic equation: 6x 2  – x – k= 0. Its one of its root: 2/ 3 If x = 2/3 is root of the given equation, then it will satisfy the given equation: Then, 6 (2/3)2 – 2/3 – k = 0 ⇒ 6 x 4/ 9 – 2/ 2 – k = 0 ⇒ 8/3 – 2/3 – k = 0 ⇒ k = 2 Hence, the value of k is 2.

Question. For what values of ‘a’, does the quadratic equation x 2  – ax + 1 = 0 not have real roots?  Solution.  Given quadratic equation is x 2  – ax + 1 = 0 On comparing the given equation with Ax 2  + Bx + C = 0, we get: A = 1, B = – a, C = 1 For real roots, D > 0 B 2  – 4AC > 0 i.e. (– a) 2  – 4 × 1 × 1 = 0 a 2  > 4 or a > √4 or – 2 > a > 2 Hence, the value of ‘a’ lies between – 2 and 2.

Question. Find the values of ‘k’ for which x = 2 is a solution of the equation kx 2  + 2x – 3 = 0. Solution.  Given equation is, kx2 + 2x – 3 = 0 if x = 2, then ⇒ k (2) 2  + 2 (2) – 3 = 0 ⇒ 4k = – 1 ⇒ k = 1/4 Hence, the value of k is =1/4

Question. If x = 3 is one root of the quadratic equation x 2  – 2kx – 6 = 0, then find the value of k.    Solution.

Question. A line segment AB of length 2m is divided at a point C into two parts such that AC2 = AB × CB. Find the length of CB.  Solution.  Let the length of AC be x The, BC = 2 – x AC2 = AB × CB (given) ∴ x 2  = 2 × (2 – x) [∵ AB = 2m (given)] ⇒ x 2  = 4 – 2x ⇒ x 2  + 2x – 4 = 0 Now, if we compare the above equation with ax 2  + bx + c = 0. Then, a = 1, b = 2, c = –4 Root of the equation are,   

Question. Find the value of k for which the roots of the quadratic equation 2x 2  + kx + 8 = 0 will have equal value. Solution.  Given: quadratic equation is 2x 2  + kx + 8 = 0 For roots of the equation to be equal. D = 0 i.e., b 2  – 4ac = 0 Here, a = 2, b = k and c = 8 ⇒ k 2  – 4 × 2 × 8 = 0 ⇒ k 2  = 64 ⇒ k = ± 8 Hence, the value of k is 8 or – 8.

Question. Show that if the roots of the following quadratic equation are equal, then ad = bc x 2 (a 2  + b 2 ) + 2(ac + bd)x + (c 2  + d 2 ) = 0 Solution.  Given, quadratic equation is : x 2 (a 2  + b 2 ) + 2(ac + bd)x + (c 2  + d 2 ) = 0 whose roots are equal. To Prove : ad = bc Proof : In the given equation, A = a 2  + b 2 B = 2(ac + bd) C = c 2  + d 2 Since, roots of the given equation are equal. ∴ Discriminant B 2  – 4AC = 0 [2(ac + bd)] 2  – 4 × (a 2  + b 2 ) (c 2  + d 2 ) = 0 ⇒ 4(ac + bd) 2 – 4(a 2 c 2  + b 2 c 2 + a 2 d 2  + b 2 d 2 ) =0 ⇒ 4(a 2 c 2  + b 2 d 2  + 2abcd) – 4a 2 c 2  – 4b 2 c 2 – 4a 2 d 2  – 4b 2 d 2  = 0 ⇒8abcd – 4a 2 d 2  – 4b 2 c 2  = 0 ⇒ a 2 .b 2  + b 2 c 2  – 2abcd = 0 ⇒ (ad – bc) 2 = 0 On taking square-root on both sides ⇒ ad – bc = 0 ⇒ ad = bc Hence, proved.

Question. For what positive values of k, does the quadratic equation 3x 2  – kx + 3 = 0 not have real roots ?  Solution.  Given: quadratic equation 3x 2  – kx + 3 = 0, has no real roots. On comparing the given equation with ax 2  + bx + c = 0, we have: a = 3, b = – k, c = 3 Then, discriminant, D = b 2  – 4ac = (– k) 2  – 4 × 3 × 3 = k 2  – 36 But for no real roots, D < O Then k 2  – 36 < 0 ⇒ k 2  < 36 ⇒ k < ± 6 ⇒ k > – 6 or k < 6 Hence, the value of k < 6 (positive value) for no real roots.

Question. Solve for x : 8x 2  – 2x – 3 = 0 Solution.  8x 2  – 2x – 3 = 0 ⇒ 8x 2  – 6x + 4x – 3 = 0 ⇒ 2x(4x – 3) + 1(4x – 3) = 0 ⇒ (4x – 3) (2x + 1) = 0 ⇒ 4x – 3 = 0 or 2x + 1 = 0 i.e., x = 3 4 or -1/ 2

Question. Solve the following quadratic equation : 6a 2 x 2  – 7abx – 3b 2  = 0  Solution.  Given: quadratic equation is : 6a 2 x 2  – 7abx – 3b 2  = 0 On comparing the given equation with AX 2  + BX + C = 0, we get: A = 6a 2 , B = –7ab, C = –3b 2 Then, discrimnants.   

Question. Solve for x : 1/ x + 4 – 1/ x – 7 = 11/ 30 , x# – 4 , 7 .  Solution.  Solve for x : 1/ x + 4 – 1/ x – 7 = 11/ 30 Now 1/ x + 4 – 1/ x – 7 = 11/ 30 ⇒ x – 7 – x – 4 / (x +4) (x -7) = 11 / 30 ⇒ (x + 4) (x – 7) + 30 = 0 ⇒ x 2  – 3x + 2 = 0 ⇒ (x – 2) (x – 1) = 0 i.e. x – 1 = 0 or x – 2 = 0 ⇒ x = 1 or 2

Question. Solve for x : √3x 2  + 10x – 8√3 = 0  Solution.  Given: 3x 2  + 10x − 8 3 = 0 On comparing the above equation with ax 2  + bx + c = 0, we get : a = √3 , b = 10 and c = – 8√3 Then, discriminant, D = b 2  – 4ac = (10) 2  – 4 × √3 × ( – 8√3 ) = 100 + 96 = 196   

Hence, the roots of the given equation are 2 √3/3 and – 4 √3.

Question. A quadratic equation with integral coefficient has integral roots. Justify your answer. Solution.  No, the given statement is not always true. Consider the quadratic equation 8x 2  – 2x – 1 = 0 By splitting the middle term, 8x 2  – 4x + 2x – 1 = 0 4x(2x – 1) + 1(2x – 1) = 0 (4x + 1)(2x – 1) = 0 If 4x + 1 = 0 ⇒ x = –1/4 2x – 1 = 0 ⇒ x = 1/2 So, the given equation has integral coefficients but no integral roots. Hence, the given statement is false.

Question. Solve for x : 6x 2  + 11x + 3 = 0 Solution.  6x 2  + 11x + 3 = 0 ⇒ 6x 2  + 9x + 2x + 3 = 0 ⇒ 3x (2x + 3) + 1 (2x + 3) = 0 ⇒ (2x + 3) (3x + 1) = 0 ⇒ 2x + 3 = 0 or 3x + 1 = 0 i.e., x = -3 / 2 or x = -1/ 3

Question. Does there exist a quadratic equation whose coefficients are rational but both of its roots are irrational ?  Solution.  Yes, there exists a quadratic equation whose coefficients are rational but both of its roots are irrational . Consider the quadratic equation x 2  – 6x + 7 = 0 Here, D = b 2  – 4ac = (–6) 2  – 4(1)(7) ⇒ D = 36 – 28 = 8 Since, discriminant is not a perfect square, therefore it will have irrational roots. The roots will be   

Question. Find the roots of the quadratic equation √2x 2  + 7x + 5 √2 = 0 .  Solution.  Given, quadratic equation is √2x 2  + 7x + 5 √2 = 0 On comparing the above equation with ax 2  + bx + x = 0 we get a = √2 , b = 7, c = 5 √2 Then, discriminant, D = b2 – 4ac = (7)2 – 4 × √2 × 5 √2 = 49 – 40 = 9 Now, x = −b ± √D/ 2a = −7 ± √9 / 2 √2 = −7 ± 3 / 2 √2 ∴ x = -4/ 2 √3 and -10/ 2√3 = -2 √3 and – 5 √3 Hence, the roots of the given equation is -2/ √3 and -5/ √3

Question. If b = 0 and c < 0, is it true that the roots of x 2  + bx + c = 0 are numerically equal and opposite in sign ?  Solution.  It is given that b = 0 and c < 0. The given quadratic equation is: x 2  + bx + c = 0 On putting b = 0 in this equation, we get x 2  + 0.x + c = 0 x 2  + c = 0 ⇒ x 2  = –c Here, c < 0 ⇒ –c > 0 ⇒ x = ± –c Hence, the roots of x 2  + bx + c = 0 are numerically equal and opposite in sign.

Question. Find the value of k for which the equation x 2  + k(2x + k – 1) + 2 = 0 has real and equal roots.  Solution.  Given, quadratic equation is: x 2  + 2xk + (k 2  – k + 2) = 0 On comparing the quadratic equation, with ax 2 + bx + c = 0, we get: a = 1, b = 2k, c = k 2  – k + 2 Since, the roots of the above equation are real and equal. ∴ Discriminant, D = 0 i.e., b 2  – 4ac = 0 (2k) 2  – 4 × 1 × (k 2  – k + 2) = 0 ⇒ 4k 2  – 4k 2  + 4k – 8 = 0 ⇒ 4k – 8 = 0 ⇒ k = 2 Hence, the value of k is 2.

Question. If x = 2/ 3 and x = – 3 are roots of the quadratic equation ax 2  + 7x + b = 0, find the values of a and b. Solution.  Since x = 2/ 3 and x = – 3 are the roots of the quadratic equation ax 2  + 7x + b = 0 Now, sum of roots: 2 /3 + (- 3 ) = – 7/ a ⇒ – 7 /3 = – 7/ a ⇒ a = 3. Product of roots: ⇒ 2/3 × (- 3 ) = b/ a ⇒ – 2 = b /3 [… a = 3] ⇒ b = – 6 Hence, the values of a and b are 3 and –6 respectively.

Question. If a and b are the roots of the equation x 2  + ax – b = 0, then find a and b.  Solution.  x 2  + ax – b = 0 Sum of the roots = a + b = Coefficient of x /Coefficient of x 2  = -a Product of roots = ab = constant term / Coefficient of x 2  = b So, a + b = –a ⇒ b = –2a and, ab = –b ⇒ a = –1 Putting the value of a, we get b = –2 × (–1) = 2 Hence, a = –1 and b = 2.

Question. Solve for x : x + 3 / x + 2 = 3x – 7/2x -3 , X  2 , 3/2 Solution.  ⇒ (x + 3) (2x – 3) = (x + 2) (3x – 7) ⇒ 2x 2  + 6x – 3x – 9 = 3x 2  + 6x – 7x – 14 ⇒ 2x 2  – 3x 2  + 3x + x – 9 + 14 = 0 ⇒ – x 2  + 4x + 5 = 0 ⇒ x 2  – 4x – 5 = 0 ⇒ x 2  – 5x + x – 5 = 0 (on splitting the middle term) ⇒ x (x – 5) + 1(x – 5) = 0 ⇒ (x + 1) (x – 5) = 0 ⇒ x = – 1, 5 Hence, the values of x are – 1 and 5.

Question. Solve for x: 2x + 9 + x = 13      Solution.  

Question. Solve the following quadratic equation for x: 4x 2  + 4bx – (a 2  – b 2 ) = 0 Solution.  4x 2  + 4bx + b 2  – a 2  = 0 ⇒ (2x + b) 2  – (a) 2  = 0 ⇒ (2x + b + a)(2x + b – a) = 0 ⇒ x = –a + b /2 + , x = a – b / 2

Question. Find the value of p, for which one root of the quadratic equation px 2  – 14x + 8 = 0 is 6 times the other.    Solution.

Question. Determine the condition for one root of the quadratic equation ax 2  + bx + c = 0 to be thrice the other.  Solution.  Let the roots of the equation ax 2  + bx + c be α and 3α. Then, sum of the roots = α + 3α = 4α = − b /a α= − b/ 4a Product of the roots = α × 3α = 3α 2  = c/a 3 (−ba/4a) = c/a 3b 2 / 16a 2  = c/a

Question. The sum of the areas of two squares is 157 m 2 . If the sum of their perimeters is 68 m, find the sides of the two squares. Solution.  Let ‘x’ and ‘y’ be the length of the sides of the two squares.the, area of first square = (side) 2  = x 2   area of second square = (side) 2  = y 2 According to the question, x 2  + y 2  = 157 Now, the perimeter of the first square = 4 × side = 4x Perimeter of the second square = 4 × side = 4y According to the question: 4x + 4y = 68 or x + y = 17 y = 17 – x …(ii) Put the value of ‘y’ from equation (ii), in equation (i). ⇒ x 2  + (17 – x)2 = 157 ⇒ x 2  + 289 + x 2 – 34x – 157 = 0 ⇒ 2x 2  – 34x + 132 = 0 ⇒ x 2  – 17x + 66 = 0 ⇒ (x – 6) (x – 11) = 0 ∴ x = 6 or 11 When, x = 6, then y = 11 when, x = 11, then y = 6 Hence, the sides of the squares are 6 m and 11m. Question. Write all the values of p for which the quadratic equation x 2  + px + 16 = 0 has equal roots. Find the roots of the equation so obtained.  Solution.  Given equation: x 2  + px + 16 = 0 Here, a = 1, b = p, c = 16 Discriminant, D = b 2  – 4ac = p 2  – 4 × 1 × 16 = p 2  – 64 If roots are equal, then: D = 0 i.e. p 2  – 64 = 0 ⇒ p 2  = 64 ⇒ p = ± 8 ∴ Equation is x 2  ± 8x + 16 = 0 ⇒ (x ± 4) 2  = 0    [∵ (a + b) 2  = a 2  ± 2ab + b 2 ] ⇒ x ± 4 = 0 ⇒ x = – 4, 4 Hence, roots are x = – 4 and x = 4 and the values of p are – 8 and 8. Question. The product of two successive integral multiples of 5 is 1050. Determine the multiples.  Solution.  Let two successive integral multiples of 5 be x and (x + 5) According to question, x(x + 5) = 1050 x 2  + 5x – 1050 = 0 (x – 30) (x + 35) = 0 x = 30 or –35 When x = 30, Multiples are 30 and 30 + 5 = 35 When x = –35, Multiples are –35 and –35 + 5 = –30

Question. If the equation (1 + m 2 )x 2  + 2 mcx + c 2  – a 2  = 0 has equal roots then show that c 2  = a 2 (1 + m 2 ). Solution.  Given: A quadratic equation is : (1 + m2)x 2  + 2 mcx + c 2  – a 2  = 0 To prove: (1 + m 2 ) x 2  + 2mcx + c 2  – a 2 = 0, with equal roots c 2  = a 2  (1 + m 2 ) we get A = 1 + m 2 , B = 2 mc, C = c 2  – a 2 The roots of the given equation is equal, then Discriminant, D = 0 ∴ B 2  – 4AC = 0 (2 mc) 2  – 4 × (1 + m 2 ) (c 2 – a 2 ) = 0 ⇒ 4m 2 c 2  – 4(c 2  + c 2 m 2  – a 2  – a 2 m 2 ) = 0 ⇒ 4m 2 c 2  – 4c 2  – 4c 2 m 2  + 4a 2 + 4m 2 a 2  = 0 ⇒ m 2 a 2  + a 2  – c 2  = 0 ⇒ c 2  = m 2 a 2  + a 2 ⇒ c 2  = a 2  (1 + m 2 ) Hence, proved Question. If the roots of the equation (a 2  + b 2 )x 2  – 2(ac + bd)x + (c 2  + d 2 ) = 0 are equal, prove that b/ a = d /c    Solution.

Question. Find the dimensions of a rectangular park whose perimeter is 60 m and area 200 m 2 .  Solution.  Let ‘l‘ be the length and ‘b’ be the breadth of the rectangular park. Perimeter of the park, p = 2(l + b) Area of the park, A = l × b According to the given conditions: 2(l + b) = 60 ⇒ l + b = 30 ⇒ l = 30 – b ….(i) and lb = 200 (30 – b)b = 200 from (i)] ⇒ 30b – b 2  = 200 b 2  – 30b – 200 = 0 on splitting the middle term, we get: ⇒ b 2  – 20b – 10b + 200 = 0 ⇒ (b – 20) (b – 10) = 0 ⇒ b = 20 or 10 when b = 20m, l = 10 m when b = 10m, l = 20 m Hence, the length and breadth of the rectangle are 10 m and 20 m or 20 m and 10 m respectively.

Question. A train travels 360 km at a uniform speed. In the speed had been 5 km/hr more, it would have taken 1 hour less for the same journey. Find the speed of the train.  Solution.  Let, the actual speed of the train be ‘x’ km/hr. Time taken by the train at actual speed, t 1  = 360 /x hr Increased speed of the train = (x + 5) km / hr Time taken by the train at the increased speed t 2  = 360/ x + 5. hr  According to the given condition: t 1  – t 2  = 1 ⇒ 360/x – 360/ x + 5 = 1 ⇒ 360 (x + 5- 5 x)/ (x + 5) x = 1 ⇒ 360 × 5 = x 2  + 5x ⇒ x 2  + 5x – 1800 = 0 ⇒ x 2  + 5x – 1800 = 0 ⇒ x 2  + 45x – 40x – 1800 = 0 ⇒ x(x + 45) – 40(x + 45) = 0 ⇒ (x – 40) (x + 45) = 0 ⇒ x = 40 [∵ x = – 45, is not possible as speed cannot be negative] Hence, the actual speed of the train is 40 km/hr.

Question. Solve for x : 1/ a + b + x = 1/a + 1/b + 1/x ;   a ≠ b ≠ 0, x ≠ 0, x ≠ – (a + b) Solution.  1/ a + b + x = 1/a + 1/b + 1/x   

⇒ – ab = ax + bx + x 2 ⇒ x 2  + ax + bx + ab = 0 ⇒ x(x + a) + b(x + a) = 0 ⇒ (x + b) (x + a) = 0 ⇒ x = – a, – b Hence, the values of x are – a and – b.

Question. Find a natural number whose square diminished by 84 is equal to thrice of 8 more than the given number.  Solution.  Let n be the required natural number. According to the question: Square of natural number diminished by 84 gives n 2 – 84. Thrice of 8 more than given number = 3(8 + n). According to the question, n 2  – 84 = 3(8 + n) ⇒ n 2  – 84 = 24 + 3n ⇒ n 2  – 3n – 108 = 0 Splitting the middle term, we have ⇒ n 2  – 12n + 9n – 108 = 0 ⇒ n(n –12) + 9(n – 12) = 0 ⇒ (n – 12)(n + 9) = 0 n = 12 or n = –9 But n ≠- 9 as n is a natural number. Hence, the required natural number is 12.

Question. If the roots of the quadratic equation (a – b)x 2  + (b – c)x + (c – a) = 0 are equal, prove that b + c = 2a.    Solution.

Question. A natural number, when increased by 12, equals 160 times its reciprocal. Find the number. Solution.  Let n be the required natural number. According to the question, number when increased by 12 is n + 12. 160 times number’s reciprocal = 160 (1/n ) = 160/n ⇒ n(n + 12) = 160 ⇒ n 2  + 12n – 160 = 0 Splitting the middle term, we have n 2  + 20n – 8n – 160 = 0 n(n + 20) – 8(n + 20) = 0 (n + 20)(n – 8) = 0 n = –20 or 8 But n ≠ -20 as n is a natural number. Hence, the required number is 8.

Question. At t minutes past 2 pm, the time needed by the minutes hand of a clock to show 3 pm was found to be 3 minutes less than t 2 /4 minutes. Find t. Solution.  It is given that at t minutes past 2 pm, the time needed by the minute hand to show 3 pm was found to be 3 minutes less than t 2 /4 min. ⇒ t + (t 2 / 4 – 3) = 60  [ ∴ time between 2 pm and 3 pm = 1 hour = 60 min.] ⇒ 4t + t 2  – 12 = 240 ⇒ t 2  + 4t – 12 – 240 = 0 ⇒ t2 + 4t – 252 = 0 Splitting the middle term, we have t 2  + 18t – 14t – 252 = 0 ⇒ t(t + 18) –14(t + 18) = 0 ⇒ (t + 18)(t – 14) = 0 t = –18 or t = 14. But t ≠ -18 as time cannot be negative ⇒ t = 14 Hence, the required value of t is 14 minutes.

Question. A plane left 30 minutes later than the scheduled time and in order to reach its destination 1500 km away on time it has to increase its speed by 250 km/hr from its usual speed. Find the usual speed of the plane.  Solution.  Let ‘x’ km / hr be the speed of the plane. Increased speed = (x + 250) km / hr   67 ⇒ x 2  + 250 x = 250 × 3000 ⇒ x 2  + 250 x – 750000 = 0 ⇒ x 2  + 1000 x – 750 x – 750000 = 0 ⇒ x (x + 1000) – 750 (x + 1000) = 0 ⇒ (x – 750) (x + 1000) = 0 ⇒ x = 750 or – 1000 ⇒ x = 750 [∵ speed cannot be negative] Hence, the usual speed of the plane is 750 km/hr.

Question. If Zeba were younger by 5 years than what she really is, then the square of her age (in years) would have been 11 more than five times her actual age. What is her age now? Solution.  Let actual age of Zeba be x years. Her age when she was 5 years younger = (x – 5) years. According to the condition given in question: Square of her age = 11 more than 5 times her actual age (x – 5) 2  = 11 + 5(x) ⇒ x2 + 25 – 10x = 11 + 5x [ a (a – b) 2 = a 2 + b 2 – 2ab] ⇒ x 2  – 10x – 5x + 25 – 11 = 0 ⇒ x – 15x + 14 = 0 Splitting the middle term, we have ⇒ x 2  – 14x – x + 14 = 0 ⇒ x(x – 14) –1(x – 14) = 0 ⇒ (x – 14)(x – 1) = 0 ⇒ x = 14 or x = 1 But x ≠ +1 as in that case (x – 5) will not be possible ⇒ x = 14 Hence, Zeba’s age now is 14 years.

Question. At present, Asha’s age (in years) is 2 more than the square of her daughter Nisha’s age. When Nisha grows to her mother’s present age, Asha’s age would be one year less than 10 times the present age of Nisha. Find the present ages of both Asha and Nisha. Solution.  Let Nisha’s present age be x years. Then, Asha’s present age = (2 + x 2 ) [By the given conditon] Now, when Nisha grows to her mother’s present age i.e. after {(x 2 + 2) – x} years. Then, Asha’s age will become {(x 2 + 2) – x} years. Now by the given condition, Asha’s age = 1 year less than 10 times present age of Nisha. (2 + x 2 ) + {(x 2  + 2) – x} = 10x – 1 ⇒ 2 + x 2  + x 2  + 2 – x = 10x – 1 ⇒ 2x 2  – 11x + 5 = 0 Splitting the middle term, we have ⇒ 2x 2  – 10x – x + 5 = 0 ⇒ 2x(x – 5) – 1(x – 5) = 0 ⇒ (x – 5)(2x – 1) = 0 ⇒ (x – 5)(2x – 1) = 0 ⇒ x = 5 or x = 1/2 But x ≠ 1/2 as then Nisha’s age = 1/2 . This means that her mother Asha’s age = (x 2  + 2) = (1/4 + 2) = 2, 1/4 years which is not possible. Hence, the present age of Nisha = 5 years and the present age of Asha = x 2  + 2 = 52 + 2 = 25 + 2 = 27 years

Question. In a class test, the sum of Arun’s marks in Hindi and English is 30. Had he got 2 marks more in Hindi and 3 marks less in English, the product of the marks would have been 210. Find his marks in the two subjects.  Solution.  Let, the Arun’s marks in Hindi be x. Then, marks in English = 30 – x According to the given condition, ⇒ (x + 2) (30 – x – 3) = 210 ⇒ (x + 2) (27 – x) = 210 ⇒ 27x + 54 – x2 – 2x = 210 ⇒ x2 – 25x + 156 = 0  x2 − 13x −12x + 156 = 0  x (x − 13) − 12(x − 13) = 0 ⇒ (x – 13) (x – 12) = 0 ⇒ x = 13 or 12 When : x = 13 Marks in Hindi = 13 Marks in English = 30 – 13 = 17 When : x = 12 Marks in Hindi = 12 Marks in English = 30 – 12 = 18 Hence, the marks obtained in the two subjects are (13, 17) or (12, 18).

Question. Solve for x : x + 3 /x – 2 -1 – x  = 17/ 4 ; 0, 2  Solution.  x + 3 /x – 2 – (1 – x)  = 17/ 4   75 ⇒ (x 2  + 3x – 3x + 2 + x 2 ) × 4 = 17(x 2  – 2x) ⇒ 4(2x 2  + 2) = 17x 2  – 34x ⇒ 8x 2  + 8 = 17×2 – 34x ⇒ 17x 2  – 8x 2  – 34x – 8 = 0 ⇒ 9x 2  – 34x – 8 = 0 ⇒ 9x 2  – 36x + 2x – 8 = 0 ⇒ 9x(x – 4) + 2(x – 4) = 0 ⇒ (9x + 2) (x – 4) = 0 ⇒ x = – 2/9  , 4 Hence, the values of x are – 2 /9 , 4

Question. A motor boat whose speed is 18 km/hr in still water takes 1 hr more to go 24 km upstream than to return downstream to the same spot. Find the speed of the stream. Solution.  Let, the speed of the stream be x km/hr. Speed of the boat in still water = 18 km/hr ∴ The speed of the boat in upstream = (18 – x) km/hr The speed of the boat in downstream = (18 + x) km/hr. Total distance to be covered = 24 km ∴ Time taken in upstream, t 1  = 24 /(18 − x) hr Time taken in downstream, t 2  = 24 (18 + x) hr According to the question, 24 /18 – x –  24/18 + x = 1 ⇒ 24 (18 + x – 18 + x) / (18 – x) (18 + x ) = 1 ⇒ 24 × 2x = 324 – x 2 ⇒ x 2  + 48x – 324 = 0 ⇒ x 2  + 54x – 6x – 324 = 0 ⇒ x(x + 54) – 6 (x + 54) = 0 ⇒ (x + 54) (x – 6) = 0 x ≠ – 54 (∵ speed can’t be negative) ∴ x = 6 Hence, the speed of the stream is 6 km/hr.

Question. A train travels at a certain average speed for a distance of 63 km and then travels at a distance of 72 km at an average speed of 6 km/hr more than its original speed. If it takes 3 hours to complete total journey, what is the original average speed ?  Solution.  Let the original speed of the train be ‘x’ km/hr. Increased speed = (x + 6) km/hr Now, time taken to cover 60 km at original speed, t 1  = 63/ x. hr Time taken to cover 72 km at increased speed: t 2  = 72 /x + 6. hr 63/x + 72/ x + 6 = 3 ⇒ 63 x + 378 + 72x / x (x + 6) = 3 ⇒ 135 x + 378 = 3 (x2 + 6x) ⇒ 3x 2  + 18x – 135x – 378 = 0 ⇒ 3x 2  – 117x – 378 = 0 ⇒ x 2  – 42x + 3x – 126 = 0 ⇒ x(x − 42) + 3(x − 42) = 0 ⇒ (x – 42) (x + 3) = 0 x = – 3 (∵ speed can’t be negative) ∴ x = 42 Hence, the original average speed of train is 42 km/hr.

Question. The altitude of a right-angled triangle is 7 cm less than its base. If the hypotenuse is 13 cm, then find the other two sides. Solution.  Let, the base of the right angled triangle be ‘x’ m. Then, the altitude of a right-angled triangle is (x – 7) cm. And, the hypotenuse of right angled triangle = 13 cm Then, by the pythagoras theorem H2 = P 2  + B 2 ⇒ 132 = (x – 7)2 + x 2 ⇒ 169 = x 2  + 49 – 14x + x 2 ⇒ 2x 2  – 14x – 120 = 0 ⇒ x 2  – 7x – 60 = 0 ⇒ x 2  – 12x + 5x – 60 = 0 ⇒ x(x – 12) + 5(x – 12) = 0 ⇒ (x + 5) (x – 12) = 0 ⇒ x = 12 (∵ x = – 5 is not possible) The base of the right angled triangle = 12 cm and altitude = 12 – 7 = 5 cm. Hence, the other two sides of triangle are 5 cm and 12 cm.

Question. Find two consecutive odd natural numbers, the sum of whose squares is 394. Solution.  Let, the first number be x and the second number be (x + 2). According to the given condition, x 2  + (x + 2) 2  = 394 ⇒ x 2  + x 2  + 4 + 4x = 394 ⇒ 2x 2  + 4x – 390 = 0 ⇒ x 2  + 2x – 195 = 0 ⇒ x 2  + 15x – 13x –195 = 0 ⇒ x(x + 15) – 13 (x + 15) = 0 ⇒ (x – 13) (x + 15) = 0 ⇒ x = 13, – 15 x ≠ –15 [∵ natural numbers are negative] Hence, the two consecutive odd natural numbers are 13 and 15.

Question. A and B working together can do a work in 6 days. If a takes 5 days less than B to finish the work, in how many days can B can do the work alone? Solution.  Let B take ‘x’ days to complete the work. Then, A takes (x – 5) days to complete the work done. According to the given condition: 1/ x + 1/ x – 5= 1 / 6 ⇒ x – 5 + x /x (x- 5) = 1/ 6 ⇒ 6(2x – 5) = x2 – 5x ⇒ x 2  – 5x – 12x + 30 = 0 ⇒ x 2  – 17x + 30 = 0 ⇒ x 2  – 15x – 2x + 30 = 0 (on splitting the middle term) ⇒ x(x – 15) – 2(x – 15) = 0 ⇒ (x − 15)(x − 2) = 0 ⇒ x = 2 or 15 But x = 2 is not possible as x < 5. ∴ x = 15 Hence, B takes 15 days to complete the work alone.

Question. Find x in terms of a, b and c : a/ x – a + b/ x – b = 2 c / x – c , x ‘ a ,b , c Solution.

Question. Solve for x :   81 2x / x – 3 + 1 /2x + 3 + 3x + 9 / (x -3) (2x + 3)  = 0 , x ≠ 3, – 3/2 Solution.

Question. Solve for x: x – 1/ 2x + 1 + 2x +1 / x -1 = 2, where x ≠-1/2 , 1.    Solution.

Question. Using quadratic formula, solve the following equation for x : 4 abx 2  + (b 2  – ac)x – bc = 0  Solution.  We have, abx 2  + (b 2  – ac) x – bc = 0 Comparing this equation with Ax 2  + Bx + C = 0, we get A = ab, B – b 2  – ac, C = -bc ∴ D = B 2  – 4AC = (b 2  – ac) 2  – 4(ab)(-bc) = b4 + a 2 c 2  – 2ab 2 c + 4ab 2 c = b 4  + 2ab 2 c + a2c 2 = (b 2  ) 2  + 2(b 2  )(ac) + (ac) 2 = (b 2  + ac) 2  > 0 So, the roots of the given equation are real and are given by :

Question. Find the roots of the quadratic equation 2x 2  – 7x – 3 = 0 by the method of completing the square. Solution.  We have, 2x 2  – 7x – 3 = 0

Question. The sum of the ages of a woman and her daughter is 40 years. The product of their ages five years ago was 125 years. Find their present ages. Solution.  Let the present age of the woman be x years. Then, the present age of her daughter is (40 – x) years. Five years ago, Age of woman = (x – 5) years and, Age of daughter = (40 – x – 5) years = (35 – x) years. According to the given question, (x – 5) (35 – x) = 125 ⇒ 35 x – x 2  – 175 + 5x – 125 = 0 ⇒ x 2  – 40 x + 300 = 0 ⇒ (x – 30) (x – 10) = 0 ⇒ x – 30 = 0 or x – 10 = 0 ⇒ x = 30 or x = 10 when, x = 30, then 40 – x = 40 – 30 = 10 Thus, when the woman’s present age is 30 years, her daughter’s present age is 10 years. When, x = 10, then 40 – x = 40 – 10 = 30, which is absurd, because the woman’s age cannot be less than her daughter’s age. Hence, present age of woman = 30 years and, present age of daughter = 10 years.

Question. Find the value(s) of k for which the following equation has equal roots. (k – 12)x 2  +2(k – 12)x+2= 0 Solution.  We have, (k -12)x 2  + 2(k -12)x + 2 = 0 Here, a = k – 12, b = 2 (k – 12), c = 2 ∴ D = b2 – 4ac = 4(k – 12)2 – 4 (k – 12) × 2 = 4 (k -12) [(k -12) – 2] = 4(k -12)(k -14) The given equation will have equal roots if D = 0. ⇒ 4 (k -12) (k -14) = 0 ⇒ k – 12 = 0 or k – 14 = 0 ⇒ k = 12 or k = 14 Ans.

Question. By increasing the list price of a book by Rs. 10 a person can buy 10 less books for Rs. 1,200. Find the original list price of the book.    Solution.  Let the original list price of book = Rs. x. and, the increased list price of a book = Rs. (x + 10)

Question. Determine if x = 3 is a root of the given equation or not :

Solution.  At x = 3

Question. A plane left 30 minutes later than the schedule time and in order to reach its destination 1500 km away in time it has to increase its speed by 250 km/hr from its usual speed. Find its usual speed. Solution.  Let the usual speed of the plane be x km/hr. Then Time taken to cover 1500 km with the usual speed = 1500 / x hrs. Time taken to cover 1500 km with the speed of (x + 250) km/hr = 1500 / x + 2500 hrs. According to given question, 1500/x – 1500/ x + 250 = 30/60 ⇒ 1500x + 1500 x 250 – 1500x / x(x + 250) = 1/2 ⇒ 1500 x 250 / x2 + 250x = 1/2 ⇒ x 2  + 250x – 750000 = 0 ⇒ x 2  + 1000x – 750 x – 750000 = 0 ⇒ x (x + 1000) – 750 (x + 1000) = 0 ⇒ (x – 750) (x + 1000) = 0 ⇒ x = 750 or x = – 1000 ⇒ x = 750 [∴ speed cannot be negative] Hence, the usual speed of the plane is 750 km/hr.

Question. A two digit number is such that the product of its digits is 18. When 63 is subtracted from the number, the digits interchange their places. Find the number.  Solution.  Let the tens digit be x. Then, the units digit = 18/x

Question. The hypotenuse of a right triangle is 3 √5 cm. If the smaller side is tripled and the larger side is doubled, the new hypotenuse will be 15 cm. Find the length of each side.  Solution.  Let the smaller side of the right triangle be x cm and the larger side be y cm. Then,

Question. Find the roots of the quadratic equation 8x 2  – 22x – 21 = 0 by factorization. Solution.  We have, 8x 2  – 22x – 21 = 0 ⇒ 8x 2  – 28x + 6x – 21 = 0 ⇒ 4x (2x – 7) + 3 (2x – 7) = 0 ⇒ (4x + 3) (2x – 7) = 0 ⇒ 4x + 3 = 0 or 2x – 7 = 0

Assignments for Class 10 Mathematics Quadratic Equation as per CBSE NCERT pattern

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Course: Class 10   >   Unit 4

• Quiz 2 Quadratic Equations

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Maths assignment class viii | quadrilateral, lesson plan math class x (ch-4) | quadratic equations.

• Method of factorizing a quadratic equations class 9 th .
• Quadratic polynomials chapter 2 class 10 th  
• Introduction explanation and definition of quadratic equations.
• Difference between the quadratic equations and quadratic polynomials.
• Relationship between roots and coefficients of quadratic polynomials.
• Method of finding the roots by factor method, by the method of completing the square and by quadratic formulas.
• Discuss the nature of roots by using discriminant.
• All formulas and all important concepts related to the quadratic equations.
• Students should know the nature of solutions by using Discriminant.
• Students should know the  factor method, method of completing the squares, and quadratic formula for solving the Q.E. 
• Students should also know the method of implementation of these formulas in simple and complex problems.

Teacher should write the quadratic equation on the board and then explain all the components of quadratic equations like coefficient, variable and constant term.

General Quadratic Equation is :  

ax 2  + bx + c = 0

a is the coefficient of x 2 ,

b is the coefficient of  x,

c is the constant term.

Now teacher will introduce the definition of quadratic equations and explain the difference between the quadratic equations and quadratic polynomials.

Solutions of the quadratic equations are called its roots.

Roots are the values of x for which the given quadratic equation become equal to zero.

Now teacher will explain the relationship between the roots and coefficients of quadratic equations.

Teacher will also explain the method of making the quadratic equation from the roots.

x 2  – Sx + P = 0

Where S is the sum of roots and P is the product of roots

x 2  – (Sum of roots)x + Product of roots = 0

Factor Method to solve Q.E.

Now teacher will introduce the factor method of finding the roots of the quadratic equations. Teacher will also provide sufficient number problems to the students so that students will understand the concept properly.

Now teacher will introduce the Discriminant(D) and explain its relation with the general Q.E.

D = b 2  – 4ac

Now teacher will explain the Nature of Roots with different conditions of  D. i.e.  for D > 0,  D = 0, D < 0  and  D ≥ 0

Now teacher will write general quadratic equation and step by step explain the method of completing the square. Teacher will also explain the method by taking 2 - 3 examples so that students will completely understand the concept.

Now teacher will explain to the students that the last result that we get in the method of completing the square is called the Quadratic formula.

Students can also solve the quadratic equations by directly applying the quadratic formula

It is important for the students to have the ability to apply all the methods of solving the quadratic equations in different situations (word problems).

Teacher will provide different situations to the students and help then to solve the problems by using different methods.

Teacher can also introduce some practice worksheets to the students.

• Students should r eview the questions given by the teacher. 
• Students should prepare the presentation on the different methods of solving quadratic equations.
• Students should Solve the  N.C.E.R.T. problems with examples, 
• Students should solve assignment on Multiple Choice Questions (MCQ) given by the teacher.

really helpful lesson plan!!!!

excellent work

Very helpful

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