assignment of quadratic equation class 10

Talk to our experts

1800-120-456-456

NCERT Solutions for Class 10 Maths Chapter 4 - Quadratic Equations

• NCERT Solutions
• Chapter 4 Quadratic Equations

Complete Resource of NCERT Class 10 Maths Chapter 4 Quadratic Equations - Free PDF Download

NCERT Solutions for Class 10 Chapter 4 Quadratic Equation , covers a crucial aspect of algebra that every student must grasp. This chapter introduces quadratic equations, detailing methods to solve them, including factoring, using the quadratic formula, and completing the square. Understanding these concepts is key, as they are foundational for higher mathematics and problem solving in science and engineering. Focus on mastering the techniques for finding the roots of quadratic equations and recognizing their practical applications. Clear explanations and step-by-step solutions in Vedantu’s materials help understand complex concepts, making them accessible and understandable.

Glance of NCERT Solutions of Maths Chapter 4 Quadratic Equations for Class 10 | Vedantu

Chapter 4 of Class 10 Maths deals with quadratic equations, which are equations  of the form  ax^2 + bx + c = 0, where a ≠ 0.

Learn about standard forms, where a, b, and c are real numbers.

The chapter focuses on finding the roots/solutions of these equations, which are the values of x that satisfy the equation.

There are different methods for solving quadratics, such as Factorization and by using Quadratic Formula

A key concept is the discriminant (b² - 4ac). It helps determine the nature of the roots:

Distinct Real Roots (D > 0): When the discriminant is positive, there are two distinct real solutions for x.

Equal Real Roots (D = 0): A positive discriminant of zero indicates two equal real roots.

No Real Roots (D < 0): A negative discriminant means there are no real number solutions, but there might be complex solutions.

The chapter also covers forming quadratic equations from word problems and applications of quadratic equations in real-life scenarios.

This article contains chapter notes important questions and Exercises link for Chapter 4 - Quadratic Equations, which you can download as PDFs.

There are four exercises and one miscellaneous exercise (24 fully solved questions) in class 10th maths chapter 4 Quadratic Equations.

Access Exercise Wise NCERT Solutions for Chapter 4 Maths Class 10

Related chapters.

Exercises under NCERT Solutions for Maths Chapter 4 Class 10 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 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. Solve the problems given in Example 1

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

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

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

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

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-

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

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

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

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

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

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

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 $\left( \text{16-x} \right)\text{years}$.

Hence, according to question-

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

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

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

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

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., $\text{y=20m}$.

Important Points from NCERT Class 10 Quadratic Equations

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.

Maths Class 10 Quadratic Equations Mind Map

Relation between the zeroes of a quadratic equation and the coefficient of a quadratic equation.

If α and β are zeroes of the quadratic equation $ax^2 + bx + c = 0$, where a, b, and c are real numbers and a ≠ 0, then

$\alpha + \beta = -\dfrac{b}{a}$

$\text{sum of zeros} = -\dfrac{\text{coefficient of x}}{\text{coefficient of }x^2}$

$\alpha \beta = \dfrac{c}{a}$

$\text{product of zeros} = -\dfrac{\text{constant term}}{\text{coefficient of }x^2}$

Methods of Solving a Quadratic Equation

The following are the methods that are used to solve quadratic equations:

(i) Factorization; (ii) Completing the Square; (iii) Quadratic Formula

Methods of Factorization

In this method, we find the roots of a quadratic equation $(ax^2 + bx + c = 0)$ by factorizing LHS into two linear factors and equating each factor to zero, e.g., $6x^2 - x - 2 = 0$ $\Rightarrow 6x^2 + 3x - 4x - 2 = 0$ …(i) $\Rightarrow 3x (2x + 1) - 2(2x + 1) = 0$ $\Rightarrow (3x - 2) (2x + 1) = 0$ $\Rightarrow 3x - 2 = 0$ or $2x + 1 = 0$

Therefore $x = \dfrac{2}{3}$ or $x = -\dfrac{-1}{2}$

Method of Completing the Square

This is the method of converting the LHS of a quadratic equation that is not a perfect square into the sum or difference of a perfect square and a constant by adding and subtracting the terms.

Quadratic Formula

Consider a quadratic equation: ax2 + bx + c = 0. If b2 – 4ac ≥ 0, then the roots of the above equation are given by:

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

Overview of Deleted Syllabus for CBSE Class 10 Maths Chapter 4 Quadratic Equation

Class 10 maths chapter 4: exercise breakdown.

NCERT Solutions for Class 10 Maths Chapter 4 - provides a comprehensive guide to understanding quadratic equations. This chapter is important for students because it teaches topics that are fundamental to advanced mathematics. The solutions describe how to solve quadratic equations, apply the quadratic formula, and investigate the nature of the roots using the discriminant. For effective exam preparation, focus on understanding formula derivation and applying the many types of problem-solving approaches described in this chapter. Last year, four to six questions from this area featured in the board exams, demonstrating its importance. These answers are precisely crafted to help students succeed by improving their problem-solving skills and conceptual understanding.

Other Study Materials of CBSE Class 10 Maths Quadratic Equation

Chapter-specific ncert solutions for class 10 maths.

Given below are the chapter-wise NCERT Solutions for Class 10 Maths . Go through these chapter-wise solutions to be thoroughly familiar with the concepts.

NCERT Study Resources for Class 10 Maths

For complete preparation of Maths for CBSE Class 10 board exams, check out the following links for different study materials available at Vedantu.

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.

• Math Article
• Quadratic Equation For Class 10

Quadratic Equation Class 10 Notes Chapter 4

Cbse class 10 maths quadratic equation notes:- download pdf here.

Get the complete concepts covered in quadratic equations for Class 10 Maths here. These quadratic equations notes help the students to recall the important definitions, formulas and tricks to solve the problems in the CBSE Board Exams 2023-24. In this article, you will learn the concept of quadratic equations, standard form, nature of roots, and methods for finding the solution for the given quadratic equations with more examples.

Introduction to Quadratic Equations

Quadratic polynomial.

A polynomial of the form a x 2 + b x + c , where a, b and c are real numbers and a ≠ 0 is called a quadratic polynomial.

For more information on Quadratic Polynomials, watch the below video

Quadratic Equation

When we equate a quadratic polynomial to a constant, we get a quadratic equation.

Any equation of the form p(x) = k, where p(x) is a polynomial of degree 2, and c is a constant, is a quadratic equation.

The Standard Form of a Quadratic Equation

The standard form of a quadratic equation is ax 2 +bx+c=0, where a,b and c are real numbers and a≠0. ‘a’ is the coefficient of x 2 . It is called the quadratic coefficient. ‘b’ is the coefficient of x. It is called the linear coefficient. ‘c’ is the constant term.

To know more about Quadratic Equations, visit here .

Solving Quadratic Equations by Factorisation

Roots of a quadratic equation.

The values of x for which a quadratic equation is satisfied are called the roots of the quadratic equation.

If α is a root of the quadratic equation ax 2 +bx+c=0, then aα 2 +bα+c=0.

A quadratic equation can have two distinct real roots, two equal roots or real roots may not exist.

Graphically, the roots of a quadratic equation are the points where the graph of the quadratic polynomial cuts the x-axis.

Consider the graph of a quadratic equation x 2 −4=0

In the above figure, -2 and 2 are the roots of the quadratic equation x 2 −4=0 Note:

• If the graph of the quadratic polynomial cuts the x-axis at two distinct points, then it has real and distinct roots.
• If the graph of the quadratic polynomial touches the x-axis, then it has real and equal roots.
• If the graph of the quadratic polynomial does not cut or touch the x-axis then it does not have any real roots.

Solving a Quadratic Equation by Factorization Method

Consider a quadratic equation 2x 2 −5x+3=0

⇒2x 2 −2x−3x+3=0

This step is splitting the middle term.

We split the middle term by finding two numbers (-2 and -3) such that their sum is equal to the coefficient of x and their product is equal to the product of the coefficient of x 2 , and the constant.

(-2) + (-3) = (-5)

And (-2) × (-3) = 6

2x 2 −2x−3x+3=0

2x(x−1)−3(x−1)=0

(x−1)(2x−3)=0

In this step, we have expressed the quadratic polynomial as a product of its factors.

Thus, x = 1 and x =3/2 are the roots of the given quadratic equation.

This method of solving a quadratic equation is called the factorisation method.

For more information on Solving a Quadratic Equation by Factorization Method, watch the below video

To know more about Solving Quadratic Equation by Factorisation, visit here .

Solving a Quadratic Equation by Completion of Squares Method

In the method of completing the squares, the quadratic equation is expressed in the form (x±k) 2 =p 2 .

Consider the quadratic equation 2x 2 −8x=10 (i) Express the quadratic equation in standard form. 2x 2 −8x−10=0

(ii) Divide the equation by the coefficient of x 2 to make the coefficient of x 2 equal to 1. x 2 −4x−5=0

(iii) Add the square of half of the coefficient of x to both sides of the equation to get an expression of the form x 2 ±2kx+k 2 . (x 2 −4x+4)−5=0+4

(iv) Isolate the above expression, (x±k) 2 on the LHS to obtain an equation of the form (x±k) 2 =p 2 (x−2) 2 =9

(v) Take the positive and negative square roots. x−2=±3

x=−1 or x=5

To know more about Solving Quadratic Equations by Completing the Square, visit here .

Solving Quadratic Equation Using Quadratic Formula

Quadratic formula.

Quadratic Formula is used to directly obtain the roots of a quadratic equation from the standard form of the equation.

For the quadratic equation ax 2 +bx+c=0,

x= [-b± √(b 2 -4ac)]/2a

By substituting the values of a,b and c, we can directly get the roots of the equation.

Example: If x 2 – 5x + 6 = 0 is the quadratic equation, find the roots.

Solution: Given, x 2 – 5x + 6 = 0 is the quadratic equation.

On comparing with the standard quadratic equation, we have;

ax 2 + bx + c = 0

a = 1, b = -5 and c = 6

b 2 – 4ac = (-5) 2 – 4 × 1 × 6 = 25 – 24 = 1 > 0 Hence, the roots are real. Using quadratic formula,

x = [-b ± √(b 2 – 4ac)]/ 2a

= [-(-5) ± √1]/ 2(1)

= [5 ± 1]/ 2

i.e. x = (5 + 1)/2 and x = (5 – 1)/2

x = 6/2, x = 4/2

Therefore, the roots of the quadratic equation are 3 and 2.

To know more about Quadratic Formula, visit here .

Discriminant

For a quadratic equation of the form ax 2 +bx+c=0, the expression b 2 −4ac is called the discriminant (denoted by D) of the quadratic equation.

The discriminant determines the nature of the roots of the quadratic equation based on the coefficients of the quadratic equation.

For more information on Discriminant, watch the below video

To know more about Discriminant Formula, visit here .

Nature of Roots

Based on the value of the discriminant, D=b 2 −4ac, the roots of a quadratic equation, ax 2 + bx + c = 0, can be of three types.

Case 1: If D>0 , the equation has two distinct real roots .

Case 2: If D=0 , the equation has two equal real roots .

Case 3: If D<0 , the equation has no real roots .

Solving using Quadratic Formula when D>0

Solve 2x 2 −7x+3=0 using the quadratic formula.

(i) Identify the coefficients of the quadratic equation. a = 2,b = -7,c = 3

(ii) Calculate the discriminant, b 2 −4ac

D=(−7) 2 −4×2×3= 49-24 = 25

D> 0, therefore, the roots are distinct.

(iii) Substitute the coefficients in the quadratic formula to find the roots

x= [-(-7)± √((-7) 2 -4(2)(3))]/2(2)

x=3 and x= 1/2 are the roots.

Solving Quadratic Equation when D=0

Let us take an example of quadratic equation 3x 2 – 2x + 1/3 = 0.

Here, a = 3, b = -2 and c = 1/3

Determinant, D = b 2 – 4ac = (-2) 2 – 4 (3)(1/3) = 4 – 4 = 0

Thus, the given equation has equal roots.

Hence the roots are -b/2a and -b/2a, i.e., 1/3 and 1/3.

Solving Quadratic Equation when D < 0

Suppose the quadratic equation is 4x 2 + 3x + 5 = 0

Comparing with the standard form of quadratic equation, ax 2 + bx + c = 0,

a = 4, b = 3, c = 5

By the formula of determinant, we know;

Determinant (D) = b 2 – 4ac

= (3) 2 – 4(4)(5)

= -71 2 – 4ac)]/ 2a

= [-3 ± √(-71)]/ 2(4)

= [-3 ± √(i 2 71)]/ 8

= (-3 ± i√71)/8

Thus, the non-real roots of the equation are x = (-3 + i√71)/8 and x (-3 – i√71)/8.

For more information on Nature Of Roots, watch the below videos

To know more about the Nature of Roots, visit here .

Graphical Representation of a Quadratic Equation

The graph of a quadratic polynomial is a parabola. The roots of a quadratic equation are the points where the parabola cuts the x-axis i.e. the points where the value of the quadratic polynomial is zero.

Now, the graph of x 2 +5x+6=0 is:

In the above figure, -2 and -3 are the roots of the quadratic equation x 2 +5x+6=0.

For a quadratic polynomial ax 2 +bx+c,

If a>0, the parabola opens upwards. If a 2 −4ac

If D>0 , the parabola cuts the x-axis at exactly two distinct points. The roots are distinct. This case is shown in the above figure in a, where the quadratic polynomial cuts the x-axis at two distinct points.

If D=0 , the parabola just touches the x-axis at one point and the rest of the parabola lies above or below the x-axis. In this case, the roots are equal. This case is shown in the above figure in b, where the quadratic polynomial touches the x-axis at only one point .

If D<0 , the parabola lies entirely above or below the x-axis and there is no point of contact with the x-axis. In this case, there are no real roots. This case is shown in the above figure in c, where the quadratic polynomial neither cuts nor touches the x-axis.

Formation of a quadratic equation from its roots

To find out the standard form of a quadratic equation when the roots are given:

Let α and β be the roots of the quadratic equation ax 2 +bx+c=0. Then,

(x−α)(x−β)=0

On expanding, we get,

x 2 −(α+β)x+αβ=0, which is the standard form of the quadratic equation.

Here, a=1,b=−(α+β) and c=αβ.

Example: Form the quadratic equation if the roots are −3 and 4.

Solution: Given -3 and 4 are the roots of the equation.

Sum of roots = -3 + 4 = 1

Product of the roots = (-3).(4) = -12

As we know, the standard form of a quadratic equation is:

x 2 − (sum of roots)x + (product of roots) = 0

Therefore, by putting the values, we get

x 2 – x – 12 = 0

Which is the required quadratic equation.

Sum and Product of Roots of a Quadratic Equation

Sum of roots = α + β =-b/a

Product of roots = αβ = c/a

Example: Given, x 2 − 5x + 8 = 0 is the quadratic equation. Find the sum and product of its roots.

Solution: x 2 − 5x + 8 = 0 is the quadratic equation given in the form of ax 2 + bx + c = 0. Hence, a = 1 b = -5 c = 8

Sum of roots = -b/a = 5

Product of roots = c/a = 8

To know more about Sum and Product of Roots of a Quadratic equation, visit here .

For more information on Sum and Product of Roots of a Quadratic equation, watch the below video

Related Links:

• NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations
• NCERT Exemplar Class 10 Maths Solutions for Chapter 4 – Quadratic Equations
• RD Sharma Solutions for Class 10 Maths Chapter 8 Quadratic Equations
• Class 10 Maths Chapter 4 Quadratic Equations MCQs
• Important Questions Class 10 Maths Chapter 4 Quadratic Equations

Practice Questions on Quadratic Equations Class 10

1. Check whether the following are quadratic equations: (i) (x – 2) 2 + 1 = 2x – 3 (ii) x(x + 1) + 8 = (x + 2) (x – 2) (iii) x (2x + 3) = x 2 + 1 (iv) (x + 2) 3 = x 3 – 4 2. Find two numbers whose sum is 27 and product is 182. 3. Find the roots of 4x 2 + 3x + 5 = 0 by the method of completing the square. 4. Find the roots of the quadratic equation 3x 2 – 5x + 2 = 0, if they exist, using the quadratic formula. 5. Find the values of k for which the quadratic equation kx(x – 2) + 6 = 0 has two equal roots.

Put your understanding of this concept to test by answering a few MCQs. Click ‘Start Quiz’ to begin!

Select the correct answer and click on the “Finish” button Check your score and answers at the end of the quiz

Visit BYJU’S for all Maths related queries and study materials

Your result is as below

Request OTP on Voice Call

Leave a Comment Cancel reply

Your Mobile number and Email id will not be published. Required fields are marked *

Post My Comment

Thankyou so much byju’s team😍😘😗😙😚☺

This information could be helped greatly for reference. Thank you so much and keep providing us with such information.

This content is very useful for my exam and it is so useful for me thank u so much Byju’s team

Awesome notes and videos it will help me to score good marks in short time, thank sirs and all team

The contents is superb and easy to understand Well done byjus team And thanku so much

Register with BYJU'S & Download Free PDFs

Register with byju's & watch live videos.

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations in Hindi and English Medium updated for final exams 2024-25. The question answers and explanation of chapter 4 of 10th Maths are based on NCERT textbooks published for 2024-2025. 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

The Hindi Medium solutions for the academic session 2024-25 for class 10 Maths chapter 4 are given here. These solutions are helpful for UP Board students studying in Class 10. 10th Maths Chapter 4 solutions are online for study 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 current 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, contact us. We will try to help you in the best possible ways.

Textbook 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.

Learning quadratic equations is very important because they have many uses in various fields. Not only are they applied in other areas of mathematics and physics, but they are also useful in real-life situations. Understanding quadratic equations helps in solving many practical problems. You can download the NCERT books for the 2024-25 session, as well as revision materials and solutions, from the links provided 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 ».

Copyright 2024 by Tiwari Academy | A step towards Free Education

NCERT Solutions Class 10 Maths Chapter 4 Quadratic Equations

NCERT solutions for class 10 maths chapter 4 Quadratic Equations deal with the concept of quadratic equations and the different ways of finding their roots. A quadratic equation is represented as ax 2 + bx + c = 0 , where a, b, c are the values of real numbers, and the value of ‘a’ is not equal to zero. This is also known as the standard form of the quadratic equation. An interesting fact to note is that many people believe that Babylonians were the first to solve quadratic equations. For instance, they knew how to find two positive numbers with a given positive sum and a given positive product, and this problem is equivalent to solving a quadratic equation. Moreover, the Greek mathematician Euclid developed a geometrical approach for finding out lengths which, in our present-day terminology, are solutions of quadratic equations The NCERT solutions class 10 maths chapter 4 Quadratic Equations teaches kids how to solve these equations by the factorization method and completing the square method. Students will come across important formulas like the quadratic formula for finding the roots of the equation.

The major takeaways from this chapter would be that for a quadratic equation two real roots will be distinct, if b 2 – 4ac > 0 ; two coincident roots will be obtained, if b 2 – 4ac = 0 ; and no roots will be there, if b 2 – 4ac < 0. The students will also explore how quadratic equations can be applied to real-life situations. The pdf file of the class 10 maths NCERT Solutions Chapter 4 Quadratic Equations in detail can be found below and also you can find some of these in the exercises given below.

• NCERT Solutions Class 10 Maths Chapter 4 Ex 4.1
• NCERT Solutions Class 10 Maths Chapter 4 Ex 4.2
• NCERT Solutions Class 10 Maths Chapter 4 Ex 4.3
• NCERT Solutions Class 10 Maths Chapter 4 Ex 4.4

NCERT Solutions for Class 10 Maths Chapter 4 PDF

The quadratic formula was discovered in 1025 AD by Sridharacharya. Since then, the quadratic equations have been widely used not just in mathematics but for real-life situations as well. This chapter explains the concept of solving quadratic equations with the help of real world practical examples so as to keep the learning process engaging and interesting. Students can have a quick glance of each section in the NCERT solutions for class 10 maths chapter 4 Quadratic Equations from the below-mentioned links :

☛ Download Class 10 Maths NCERT Solutions Chapter 4 Quadratic Equations

NCERT Class 10 Maths Chapter 4   Download PDF

NCERT Solutions for Class 10 Maths Chapter 4 Quadratic Equations

Owing to the applicability of quadratic equations in various real-life problems, it becomes an important topic that needs to be understood well. The chapter showcases enough solved examples to ensure the coverage of the topic properly. Students will learn the basics of the various methods involved in finding the roots of quadratic equations along with the graphical representation of the equation. A brief exercise-wise analysis of NCERT Solutions Class 10 Maths Chapter 4 Quadratic Equations can be seen below :

• Class 10 Maths Chapter 4 Ex 4.1 - 2 Questions
• Class 10 Maths Chapter 4 Ex 4.2 - 6 Questions
• Class 10 Maths Chapter 4 Ex 4.3 - 11 Questions
• Class 10 Maths Chapter 4 Ex 4.4 - 5 Questions

Topics Covered: The topics covered in the class 10 maths NCERT Solutions Chapter 4 Quadratic Equations are the definition of quadratic equations , standard form of a quadratic equation, nature of roots , the concept of discriminant, quadratic formula, solution of a quadratic equation by the factorization method , and completing the square method.

Total Questions: Class 10 maths chapter 4 Quadratic Equations consists of a total of 24 questions, out of which 15 are straightforward, 5 are intermediate level questions, and 4 are difficult problems. These questions are explained in a step-wise manner. The important points are written in lucid language to encourage better comprehension.

List of Formulas in NCERT Solutions Class 10 Maths Chapter 4

NCERT solutions for class 10 maths chapter 4 will help students deal with different methods of solving a quadratic equation, and these methods involve the use of various formulas, which have been explained step by step in the chapter. Along with the formulas, kids also need to understand their implications. For example, as mentioned above the value of the discriminant can help us gain important insights into the practical applications of that particular quadratic equation. It is a good idea to make a formula sheet that will give kids a quick overview of these points as and when required. A few important formulas are as given below :

• Standard form of a quadratic equation: ax² + bx + c = 0, a ≠ 0
• Quadratic formula : [-b±√(b²-4ac)]/(2a) to find the solution of a quadratic equation.
• Discriminant : b 2 – 4ac

Important Questions for Class 10 Maths NCERT Solutions Chapter 4

Video solutions for class 10 maths ncert chapter 4, faqs on ncert solutions class 10 maths chapter 4 quadratic equations, why are ncert solutions class 10 maths chapter 4 vital for scoring well.

The NCERT solutions class 10 maths chapter 4 has been designed by scholars in their respective fields. After a lot of research, they have compiled the content into this textbook, making it a precious resource for study. The quadratic equations have widespread applications in our daily lives; hence, to understand them properly, it is advisable for the students to go through this chapter thoroughly. Also, the CBSE board recommends following these solutions, thereby making them of utmost importance.

Do I Need to Practice all Questions Given in Class 10 Maths NCERT Solutions Quadratic Equations?

The NCERT Solutions Class 10 Maths Quadratic Equations covers a wide range of topics that are important from the board exam's perspective. Problems based on topics such as the roots of a quadratic equation and the factorization method are often asked in exams. Hence, it is worthwhile to go through the entire theory and the solved examples from this chapter to get a better grasp of the concepts.

What are the Important Topics Covered in NCERT Solutions Class 10 Maths Chapter 4?

The important topics covered in the NCERT Solutions Class 10 Maths Chapter 4 are how to represent the given problem statements mathematically, what is the standard form of a quadratic equation, how to solve quadratic equations by factoring and completing the squares which is an essential topic requiring regular practice. Solving questions related to these topics will help the students score well in their board exams.

How Many Questions are there in NCERT Solutions Class 10 Maths Chapter 4?

The NCERT Solutions Class 10 Maths Chapter 4 contains a total of 24 well-researched questions by the subject experts. These 24 problems include both theoretical as well as graph-based questions. All the solutions are elaborated well in the NCERT solutions and are self-explanatory. The questions are solved in more than one method to strengthen the understanding of basic quadratic equation concepts.

How CBSE Students can utilize NCERT Solutions Class 10 Maths Chapter 4 effectively?

Students can utilize the NCERT Solutions Class 10 Maths Chapter 4 effectively by reviewing the principles and theorems explained in each section of the lesson on a frequent basis. They must then solve the exercise questions after practicing all of the examples and reviewing fundamental formulas related to factorization as well as completion of squares method of quadratic equations. This will help them build up their problem-solving approach thereby building confidence in maths.

Why Should I Practice NCERT Solutions Class 10 Maths Chapter 4?

Quadratic equations is one of the topics that is not just important for mathematics but also plays a fundamental role in a lot of real-life scenarios. If you have to calculate the length and breadth of a garden, you can use the quadratic equation. Based on this information, you can plan the quantity of a grass carpet needed for the garden. Quadratic equations are also used in fields such as astronomy, science, architecture. Owing to its vast range of applications, practicing the NCERT Solutions Class 10 Maths Chapter 4 in detail becomes very important for the students. Thus, to perfect this lesson practice is key.