unit 4 homework 6 solving quadratics by completing the square

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9.2: Completing the Square

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• Page ID 18383

Learning Objectives

• Solve quadratic equations by completing the square.

Completing the Square

In this section, we will devise a method for rewriting any quadratic equation of the form

\[a x^{2}+b x+c=0\]

in the form

\[(x-p)^{2}=q\]

This process is called completing the square. As we have seen, quadratic equations in this form can easily be solved by extracting roots. We begin by examining perfect square trinomials:

\(\begin{aligned}(x+3)^{2}=\:\:\:x^{2}\:\:\:+\:\:\:& 6 x\:\:\:+&9 \\ & \color{Cerulean}{\downarrow} &\color{Cerulean}{\uparrow} \\ & \left(\frac{6}{2}\right)^{2}=(3)^{2}=&9 \end{aligned}\)

The last term, 9, is the square of one-half of the coefficient of x . In general, this is true for any perfect square trinomial of the form \(x^{2}+bx+c\).

\(\begin{aligned}\left(x+\frac{b}{2}\right)^{2} &=x^{2}+2 \cdot \frac{b}{2} x+\left(\frac{b}{2}\right)^{2} \\ &=x^{2}+b x+\left(\frac{b}{2}\right)^{2} \end{aligned}\)

In other words, any trinomial of the form \(x^{2}+bx+c\) will be a perfect square trinomial if

\[c=\left(\frac{b}{2}\right)^{2}\]

It is important to point out that the leading coefficient must be equal to 1 for this to be true.

Example \(\PageIndex{1}\)

Complete the square \(x^{2}+8x+   ?  =(x+  ? )^{2}\)

In this example, the coefficient of the middle term b = 8, so find the value that completes the square as follows:

\(\left(\frac{b}{2}\right)^{2}=\left(\frac{8}{2}\right)^{2}=(4)^{2}=\color{Cerulean}{16}\)

The value that completes the square is 16.

\(\begin{aligned} x^{2}+8 x+16 &=(x+4)(x+4) \\ &=(x+4)^{2} \end{aligned}\)

\(x^{2}+8x+16=(x+4)^{2}\)

Example \(\PageIndex{2}\)

Complete the square \(x^{2}+3x+   ?  =(x+  ? )^{2}\)

Here b = 3, so find the value that will complete the square as follows:

\(\left(\frac{b}{2}\right)^{2}=\left(\frac{3}{2}\right)^{2}=\color{Cerulean}{\frac{9}{4}}\)

The value 9/4 completes the square:

\(\begin{aligned} x^{2}+3 x+\color{Cerulean}{\frac{9}{4}} &\color{black}{=}\left(x+\frac{3}{2}\right)\left(x+\frac{3}{2}\right) \\ &=\left(x+\frac{3}{2}\right)^{2} \end{aligned}\)

\(x^{2}+3x+94=(x+\frac{3}{2})^{2}\)

We can use this technique to solve quadratic equations. The idea is to take any quadratic equation in standard form and complete the square so that we can solve it by extracting roots. The following are general steps for solving a quadratic equation with a leading coefficient of 1 in standard form by completing the square.

Example \(\PageIndex{3}\)

Solve by completing the square: \(x^{2}+14x+46=0\).

Step 1 : Add or subtract the constant term to obtain the equation in the form \(x^{2}+bx   =c\). In this example, subtract 46 to move it to the right side of the equation.

Step 2 : Use \(\left(\frac{b}{2}\right)^{2}\) to determine the value that completes the square. Here b = 14:

\(\left(\frac{b}{2}\right)^{2}=\left(\frac{14}{2}\right)^{2}=(7)^{2}=\color{Cerulean}{49}\)

Step 3 : Add \(\left(\frac{b}{2}\right)^{2}\) to both sides of the equation and complete the square.

Step 4 : Solve by extracting roots.

The solutions are \(−7−\sqrt{3}\) or \(−7+\sqrt{3}\). The check is optional.

Example \(\PageIndex{4}\)

Solve by completing the square: \(x^{2}-18x+72=0\)

Begin by subtracting 72 from both sides.

Next, find the value that completes the square using b = −18.

\(\left(\frac{b}{2}\right)^{2}=\left(\frac{-18}{2}\right)^{2}=(-9)^{2}=\color{Cerulean}{81}\)

To complete the square, add 81 to both sides, complete the square, and then solve by extracting the roots.

At this point, separate the “plus or minus” into two equations and solve each.

\(\begin{array}{ll}{x=9-3} & {\text { or } \quad x=9+3} \\ {x=6} & \quad\quad\:\:{x=12}\end{array}\)

The solutions are 6 and 12.

Note that in the previous example the solutions are integers. If this is the case, then the original equation will factor.

If it factors, we can solve it by factoring. However, not all quadratic equations will factor.

Example \(\PageIndex{5}\)

Solve by completing the square: \(x^{2}+10x+1=0\).

Begin by subtracting 1 from both sides of the equation.

Here b = 10, and we determine the value that completes the square as follows:

\(\left(\frac{b}{2}\right)^{2}=\left(\frac{10}{2}\right)^{2}=(5)^{2}=\color{Cerulean}{25}\)

To complete the square, add 25 to both sides of the equation.

\(\begin{array}{l}{x^{2}+10 x=-1} \\ {x^{2}+10 x\color{Cerulean}{+25}\color{black}{=}-1\color{Cerulean}{+25}} \\ {x^{2}+10 x\color{Cerulean}{+25}\color{black}{=}24}\end{array}\)

Factor and then solve by extracting roots.

The solutions are \(-5 - 2 \sqrt{6}\) and \(-5 + 2 \sqrt{6}\)

Sometimes quadratic equations do not have real solutions.

Example \(\PageIndex{6}\)

Solve by completing the square: \(x^{2}−2x+3=0\).

Begin by subtracting 3 from both sides of the equation.

Here b = −2, and we have

\(\left(\frac{b}{2}\right)^{2}=\left(\frac{-2}{2}\right)^{2}=(-1)^{2}=\color{Cerulean}{1}\)

At this point we see that extracting the root leads to the square root of a negative number.

No real solution

Exercise \(\PageIndex{1}\)

Solve by completing the square: \(x^{2}−2x−27=0\).

\(x=1\)±\(2\sqrt{7}\)

The coefficient of x is not always divisible by 2.

Example \(\PageIndex{7}\)

Solve by completing the square: \(x^{2}+3x−2=0\)

Begin by adding 2 to both sides.

\(\begin{array}{l}{x^{2}+3 x-2=0} \\ {x^{2}+3 x=2}\end{array}\)

Use b = 3 to find the value that completes the square:

\(\left(\frac{3}{2}\right)^{2}=\color{Cerulean}{\frac{9}{4}}\)

To complete the square, add 9/4 to both sides of the equation.

\(\begin{array}{c}{x^{2}+3 x=2} \\ {x^{2}+3 x\color{Cerulean}{+\frac{9}{4}}\color{black}{=}2\color{Cerulean}{+\frac{9}{4}}} \\ {\left(x+\frac{3}{2}\right)\left(x+\frac{3}{2}\right)=\frac{8}{4}+\frac{9}{4}} \\ {\left(x+\frac{3}{2}\right)^{2}=\frac{17}{4}}\end{array}\)

Solve by extracting roots.

The solutions are \(\frac{-3 \pm \sqrt{17}}{2}\)

So far, all of the examples have had a leading coefficient of 1. The formula \((\frac{b}{2})^{2}\) determines the value that completes the square only if the leading coefficient is 1. If this is not the case, then simply divide both sides by the leading coefficient.

Example \(\PageIndex{8}\)

Solve by completing the square: \(2x^{2}+5x-1=0\).

Notice that the leading coefficient is 2. Therefore, divide both sides by 2 before beginning the steps required to solve by completing the square.

\(\begin{array}{l}{\frac{2 x^{2}+5 x-1}{\color{Cerulean}{2}}\color{black}{=}\frac{0}{\color{Cerulean}{2}}} \\ {\frac{2 x^{2}}{2}+\frac{5 x}{2}-\frac{1}{2}=0} \\ {x^{2}+\frac{5}{2} x-\frac{1}{2}=0}\end{array}\)

Begin by adding 1/2 to both sides of the equation.

\(\begin{array}{l}{x^{2}+\frac{5}{2} x-\frac{1}{2}=0} \\ {x^{2}+\frac{5}{2} x=\frac{1}{2}}\end{array}\)

Here b = 5/2, and we can find the value that completes the square as follows:

\(\left(\frac{b}{2}\right)^{2}=\color{black}{\left(\frac{\color{Cerulean}{\frac{5}{2}}}{2}\right)}^{2}=\left(\frac{5}{2} \cdot \frac{1}{2}\right)^{2}=\left(\frac{5}{4}\right)^{2}=\color{Cerulean}{\frac{25}{16}}\)

To complete the square, add 25/16 to both sides of the equation.

\(\begin{aligned} x^{2}+\frac{5}{2} x &=\frac{1}{2} \\ x^{2}+\frac{5}{2} x\color{Cerulean}{+\frac{25}{16}} &\color{black}{=}\frac{1}{2}\color{Cerulean}{+\frac{25}{16}} \\\left(x+\frac{5}{4}\right)\left(x+\frac{5}{4}\right) &=\frac{8}{16}+\frac{25}{16} \\\left(x+\frac{5}{4}\right)^{2} &=\frac{33}{16} \end{aligned}\)

Next, solve by extracting roots.

The solutions are \(\frac{-5 \pm \sqrt{33}}{4}\)

Exercise \(\PageIndex{2}\)

Solve \(2x^{2}-2x-3=0\).

\(\frac{1\pm\sqrt{13}}{4}\)

Key Takeaways

• Solve any quadratic equation by completing the square.
• You can apply the square root property to solve an equation if you can first convert the equation to the form \((x−p)^{2}=q\).
• To complete the square, first make sure the equation is in the form \(x^{2}+bx   =c\). Then add the value \((\frac{b}{2})^{2}\) to both sides and factor.
• The process for completing the square always works, but it may lead to some tedious calculations with fractions. This is the case when the middle term, b , is not divisible by 2.

Exercise \(\PageIndex{3}\) completing the square

Complete the square.

• \(x^{2}+6x+   ?  =(x+  ? )^{2}\)
• \(x^{2}+8x+   ?  =(x+  ? )^{2}\)
• \(x^{2}−2x+   ?  =(x−  ? )^{2}\)
• \(x^{2}−4x+   ?  =(x−  ? )^{2}\)
• \(x^{2}+7x+   ?  =(x+  ? )^{2}\)
• \(x^{2}+3x+   ?  =(x+  ? )^{2}\)
• \(x^{2}+23x+   ?  =(x+  ? )^{2}\)
• \(x^{2}+45x+   ?  =(x+  ? )^{2}\)
• \(x^{2}+34x+   ?  =(x+  ? )^{2}\)
• \(x^{2}+53x+   ?  =(x+  ? )^{2}\)

1. \(x^{2}+6x+9=(x+ 3)^{2}\)

3. \(x^{2}−2x+1=(x− 1)^{2}\)

5. \(x^{2}+7x+494=(x+ 72)^{2}\)

7. \(x^{2}+23x+19=(x+ 13)^{2}\)

9. \(x^{2}+34x+964=(x+ 38)^{2}\)

Exercise \(\PageIndex{4}\)

Solve by factoring and then solve by completing the square. Check answers.

• \(x^{2}+2x−8=0\)
• \(x^{2}−8x+15=0\)
• \(y^{2}+2y−24=0\)
• \(y^{2}−12y+11=0\)
• \(t^{2}+3t−28=0\)
• \(t^{2}−7t+10=0\)
• \(2x^{2}+3x−2=0\)
• \(3x^{2}−x−2=0\)
• \(2y^{2}−y−1=0\)
• \(2y^{2}+7y−4=0\)

1. −4, 2

3. −6, 4

5. −7, 4

7. 1/2, −2

9. −1/2, 1

Exercise \(\PageIndex{5}\)

Solve by completing the square.

• \(x^{2}+6x−1=0\)
• \(x^{2}+8x+10=0\)
• \(x^{2}−2x−7=0\)
• \(x^{2}−6x−3=0\)
• \(x^{2}−2x+4=0\)
• \(x^{2}−4x+9=0\)
• \(t^{2}+10t−75=0\)
• \(t^{2}+12t−108=0\)
• \(x^{2}−4x−1=15\)
• \(x^{2}−12x+8=−10\)
• \(y^{2}−20y=−25\)
• \(y^{2}+18y=−53\)
• \(x^{2}−0.6x−0.27=0\)
• \(x^{2}−1.6x−0.8=0\)
• \(x^{2}−23x−13=0\)
• \(x^{2}−45x−15=0\)
• \(x^{2}+x−1=0\)
• \(x^{2}+x−3=0\)
• \(y^{2}+3y−2=0\)
• \(y^{2}+5y−3=0\)
• \(x^{2}+3x+5=0\)
• \(x^{2}+x+1=0\)
• \(x^{2}−7x+112=0\)
• \(x^{2}−9x+32=0\)
• \(t^{2}−12t−1=0\)
• \(t^{2}−13t−2=0\)
• \(x^{2}−1.7x−0.0875=0\)
• \(x^{2}+3.3x−1.2775=0\)
• \(4x^{2}−8x−1=0\)
• \(2x^{2}−4x−3=0\)
• \(3x^{2}+6x+1=0\)
• \(5x^{2}+10x+2=0\)
• \(3x^{2}+2x−3=0\)
• \(5x^{2}+2x−5=0\)
• \(4x^{2}−12x−15=0\)
• \(2x^{2}+4x−43=0\)
• \(2x^{2}−4x+10=0\)
• \(6x^{2}−24x+42=0\)
• \(2x^{2}−x−2=0\)
• \(2x^{2}+3x−1=0\)
• \(3x^{2}+2x−2=0\)
• \(3x^{2}−x−1=0\)
• \(x(x+1)−11(x−2)=0\)
• \((x+1)(x+7)−4(3x+2)=0\)
• \(y^{2}=(2y+3)(y−1)−2(y−1)\)
• \((2y+5)(y−5)−y(y−8)=−24\)
• \((t+2)^{2}=3(3t+1)\)
• \((3t+2)(t−4)−(t−8)=1−10t\)

1. \(−3\pm\sqrt{10}\)

3. \(1\pm 2\sqrt{2}\)

5. No real solution

7. −15, 5

9. \(2(1\pm\sqrt{5})\)

11. \(5(2\pm \sqrt{3})\)

13. −0.3, 0.9

15. −1/3, 1

17. \(\frac{-1\pm \sqrt{5}}{2}\)

19. \(\frac{-3\pm \sqrt{17}}{2}\)

21. No real solution

23. \(=\frac{7}{2}\pm i \frac{\sqrt{399}}{2}\)

25. \(6\pm\sqrt{37}\)

27. −0.05, 1.75

29. \(\frac{2\pm \sqrt{5}}{2}\)

31. \(\frac{-3\pm \sqrt{6}}{3}\)

33. \(\frac{-1\pm \sqrt{10}}{3}\)

35. \(\frac{3\pm 2 \sqrt{6}}{2}\)

37. No real solution

39. \(\frac{1\pm \sqrt{17}}{4}\)

41. \(\frac{-1\pm \sqrt{7}}{3}\)

43. \(5\pm\sqrt{3}\)

45. \(1\pm 5\sqrt{2}\)

47. \(\frac{5\pm \sqrt{21}}{2}\)

Exercise \(\PageIndex{6}\)

Solve by completing the square and round off the solutions to the nearest hundredth

• \((2x−1)^{2}=2x\)
• \((3x−2)^{2}=5−15x\)
• \((2x+1)(3x+1)=9x+4\)
• \((3x+1)(4x−1)=17x−4\)
• \(9x(x−1)−2(2x−1)=−4x\)
• \((6x+1)2−6(6x+1)=0\)

1. 0.19, 1.31

3. −0.45, 1.12

5. 0.33, 0.67

Exercise \(\PageIndex{7}\) discussion board

• Research and discuss the Hindu method for completing the square.
• Explain why the technique for completing the square described in this section requires that the leading coefficient be equal to 1.

1. Answers may vary

9.2 Solve Quadratic Equations by Completing the Square

Learning objectives.

By the end of this section, you will be able to:

• Complete the square of a binomial expression
• Solve quadratic equations of the form x 2 + b x + c = 0 x 2 + b x + c = 0 by completing the square
• Solve quadratic equations of the form a x 2 + b x + c = 0 a x 2 + b x + c = 0 by completing the square

Be Prepared 9.4

Before you get started, take this readiness quiz.

Expand: ( x + 9 ) 2 . ( x + 9 ) 2 . If you missed this problem, review Example 5.32 .

Be Prepared 9.5

Factor y 2 − 14 y + 49 . y 2 − 14 y + 49 . If you missed this problem, review Example 6.9 .

Be Prepared 9.6

Factor 5 n 2 + 40 n + 80 . 5 n 2 + 40 n + 80 . If you missed this problem, review Example 6.14 .

So far we have solved quadratic equations by factoring and using the Square Root Property. In this section, we will solve quadratic equations by a process called completing the square , which is important for our work on conics later.

Complete the Square of a Binomial Expression

In the last section, we were able to use the Square Root Property to solve the equation ( y − 7) 2 = 12 because the left side was a perfect square.

We also solved an equation in which the left side was a perfect square trinomial, but we had to rewrite it the form ( x − k ) 2 ( x − k ) 2 in order to use the Square Root Property.

What happens if the variable is not part of a perfect square? Can we use algebra to make a perfect square?

Let’s look at two examples to help us recognize the patterns.

We restate the patterns here for reference.

Binomial Squares Pattern

If a and b are real numbers,

We can use this pattern to “make” a perfect square.

We will start with the expression x 2 + 6 x . Since there is a plus sign between the two terms, we will use the ( a + b ) 2 pattern, a 2 + 2 ab + b 2 = ( a + b ) 2 .

We ultimately need to find the last term of this trinomial that will make it a perfect square trinomial. To do that we will need to find b . But first we start with determining a . Notice that the first term of x 2 + 6 x is a square, x 2 . This tells us that a = x .

What number, b, when multiplied with 2 x gives 6 x ? It would have to be 3, which is 1 2 ( 6 ) . 1 2 ( 6 ) . So b = 3.

Now to complete the perfect square trinomial, we will find the last term by squaring b , which is 3 2 = 9.

We can now factor.

So we found that adding 9 to x 2 + 6 x ‘completes the square’, and we write it as ( x + 3) 2 .

Complete a square of x 2 + b x . x 2 + b x .

• Step 1. Identify b , the coefficient of x .
• Step 2. Find ( 1 2 b ) 2 , ( 1 2 b ) 2 , the number to complete the square.
• Step 3. Add the ( 1 2 b ) 2 ( 1 2 b ) 2 to x 2 + bx .
• Step 4. Factor the perfect square trinomial, writing it as a binomial squared.

Example 9.11

Complete the square to make a perfect square trinomial. Then write the result as a binomial squared.

ⓐ x 2 − 26 x x 2 − 26 x ⓑ y 2 − 9 y y 2 − 9 y ⓒ n 2 + 1 2 n n 2 + 1 2 n

Try It 9.21

ⓐ a 2 − 20 a a 2 − 20 a ⓑ m 2 − 5 m m 2 − 5 m ⓒ p 2 + 1 4 p p 2 + 1 4 p

Try It 9.22

ⓐ b 2 − 4 b b 2 − 4 b ⓑ n 2 + 13 n n 2 + 13 n ⓒ q 2 − 2 3 q q 2 − 2 3 q

Solve Quadratic Equations of the Form x 2 + bx + c = 0 by Completing the Square

In solving equations, we must always do the same thing to both sides of the equation. This is true, of course, when we solve a quadratic equation by completing the square too. When we add a term to one side of the equation to make a perfect square trinomial, we must also add the same term to the other side of the equation.

For example, if we start with the equation x 2 + 6 x = 40, and we want to complete the square on the left, we will add 9 to both sides of the equation.

Now the equation is in the form to solve using the Square Root Property ! Completing the square is a way to transform an equation into the form we need to be able to use the Square Root Property.

Example 9.12

How to solve a quadratic equation of the form x 2 + b x + c = 0 x 2 + b x + c = 0 by completing the square.

Solve by completing the square: x 2 + 8 x = 48 . x 2 + 8 x = 48 .

Try It 9.23

Solve by completing the square: x 2 + 4 x = 5 . x 2 + 4 x = 5 .

Try It 9.24

Solve by completing the square: y 2 − 10 y = −9 . y 2 − 10 y = −9 .

The steps to solve a quadratic equation by completing the square are listed here.

Solve a quadratic equation of the form x 2 + b x + c = 0 x 2 + b x + c = 0 by completing the square.

• Step 1. Isolate the variable terms on one side and the constant terms on the other.
• Step 2. Find ( 1 2 · b ) 2 , ( 1 2 · b ) 2 , the number needed to complete the square. Add it to both sides of the equation.
• Step 3. Factor the perfect square trinomial, writing it as a binomial squared on the left and simplify by adding the terms on the right
• Step 4. Use the Square Root Property.
• Step 5. Simplify the radical and then solve the two resulting equations.
• Step 6. Check the solutions.

When we solve an equation by completing the square, the answers will not always be integers.

Example 9.13

Solve by completing the square: x 2 + 4 x = −21 . x 2 + 4 x = −21 .

Try It 9.25

Solve by completing the square: y 2 − 10 y = −35 . y 2 − 10 y = −35 .

Try It 9.26

Solve by completing the square: z 2 + 8 z = −19 . z 2 + 8 z = −19 .

In the previous example, our solutions were complex numbers. In the next example, the solutions will be irrational numbers.

Example 9.14

Solve by completing the square: y 2 − 18 y = −6 . y 2 − 18 y = −6 .

Another way to check this would be to use a calculator. Evaluate y 2 − 18 y y 2 − 18 y for both of the solutions. The answer should be −6 . −6 .

Try It 9.27

Solve by completing the square: x 2 − 16 x = −16 . x 2 − 16 x = −16 .

Try It 9.28

Solve by completing the square: y 2 + 8 y = 11 . y 2 + 8 y = 11 .

We will start the next example by isolating the variable terms on the left side of the equation.

Example 9.15

Solve by completing the square: x 2 + 10 x + 4 = 15 . x 2 + 10 x + 4 = 15 .

Try It 9.29

Solve by completing the square: a 2 + 4 a + 9 = 30 . a 2 + 4 a + 9 = 30 .

Try It 9.30

Solve by completing the square: b 2 + 8 b − 4 = 16 . b 2 + 8 b − 4 = 16 .

To solve the next equation, we must first collect all the variable terms on the left side of the equation. Then we proceed as we did in the previous examples.

Example 9.16

Solve by completing the square: n 2 = 3 n + 11 . n 2 = 3 n + 11 .

Try It 9.31

Solve by completing the square: p 2 = 5 p + 9 . p 2 = 5 p + 9 .

Try It 9.32

Solve by completing the square: q 2 = 7 q − 3 . q 2 = 7 q − 3 .

Notice that the left side of the next equation is in factored form. But the right side is not zero. So, we cannot use the Zero Product Property since it says “If a · b = 0 , a · b = 0 , then a = 0 or b = 0.” Instead, we multiply the factors and then put the equation into standard form to solve by completing the square.

Example 9.17

Solve by completing the square: ( x − 3 ) ( x + 5 ) = 9 . ( x − 3 ) ( x + 5 ) = 9 .

Try It 9.33

Solve by completing the square: ( c − 2 ) ( c + 8 ) = 11 . ( c − 2 ) ( c + 8 ) = 11 .

Try It 9.34

Solve by completing the square: ( d − 7 ) ( d + 3 ) = 56 . ( d − 7 ) ( d + 3 ) = 56 .

Solve Quadratic Equations of the Form ax 2 + bx + c = 0 by Completing the Square

The process of completing the square works best when the coefficient of x 2 is 1, so the left side of the equation is of the form x 2 + bx + c . If the x 2 term has a coefficient other than 1, we take some preliminary steps to make the coefficient equal to 1.

Sometimes the coefficient can be factored from all three terms of the trinomial. This will be our strategy in the next example.

Example 9.18

Solve by completing the square: 3 x 2 − 12 x − 15 = 0 . 3 x 2 − 12 x − 15 = 0 .

To complete the square, we need the coefficient of x 2 x 2 to be one. If we factor out the coefficient of x 2 x 2 as a common factor, we can continue with solving the equation by completing the square.

Try It 9.35

Solve by completing the square: 2 m 2 + 16 m + 14 = 0 . 2 m 2 + 16 m + 14 = 0 .

Try It 9.36

Solve by completing the square: 4 n 2 − 24 n − 56 = 8 . 4 n 2 − 24 n − 56 = 8 .

To complete the square, the coefficient of the x 2 must be 1. When the leading coefficient is not a factor of all the terms, we will divide both sides of the equation by the leading coefficient! This will give us a fraction for the second coefficient. We have already seen how to complete the square with fractions in this section.

Example 9.19

Solve by completing the square: 2 x 2 − 3 x = 20 . 2 x 2 − 3 x = 20 .

To complete the square we need the coefficient of x 2 x 2 to be one. We will divide both sides of the equation by the coefficient of x 2 . Then we can continue with solving the equation by completing the square.

Try It 9.37

Solve by completing the square: 3 r 2 − 2 r = 21 . 3 r 2 − 2 r = 21 .

Try It 9.38

Solve by completing the square: 4 t 2 + 2 t = 20 . 4 t 2 + 2 t = 20 .

Now that we have seen that the coefficient of x 2 must be 1 for us to complete the square, we update our procedure for solving a quadratic equation by completing the square to include equations of the form ax 2 + bx + c = 0.

Solve a quadratic equation of the form a x 2 + b x + c = 0 a x 2 + b x + c = 0 by completing the square.

• Step 1. Divide by a a to make the coefficient of x 2 term 1.
• Step 2. Isolate the variable terms on one side and the constant terms on the other.
• Step 3. Find ( 1 2 · b ) 2 , ( 1 2 · b ) 2 , the number needed to complete the square. Add it to both sides of the equation.
• Step 4. Factor the perfect square trinomial, writing it as a binomial squared on the left and simplify by adding the terms on the right
• Step 5. Use the Square Root Property.
• Step 6. Simplify the radical and then solve the two resulting equations.
• Step 7. Check the solutions.

Example 9.20

Solve by completing the square: 3 x 2 + 2 x = 4 . 3 x 2 + 2 x = 4 .

Again, our first step will be to make the coefficient of x 2 one. By dividing both sides of the equation by the coefficient of x 2 , we can then continue with solving the equation by completing the square.

Try It 9.39

Solve by completing the square: 4 x 2 + 3 x = 2 . 4 x 2 + 3 x = 2 .

Try It 9.40

Solve by completing the square: 3 y 2 − 10 y = −5 . 3 y 2 − 10 y = −5 .

Access these online resources for additional instruction and practice with completing the square.

• Completing Perfect Square Trinomials
• Completing the Square 1
• Completing the Square to Solve Quadratic Equations
• Completing the Square to Solve Quadratic Equations: More Examples
• Completing the Square 4

Section 9.2 Exercises

Practice makes perfect.

In the following exercises, complete the square to make a perfect square trinomial. Then write the result as a binomial squared.

ⓐ m 2 − 24 m m 2 − 24 m ⓑ x 2 − 11 x x 2 − 11 x ⓒ p 2 − 1 3 p p 2 − 1 3 p

ⓐ n 2 − 16 n n 2 − 16 n ⓑ y 2 + 15 y y 2 + 15 y ⓒ q 2 + 3 4 q q 2 + 3 4 q

ⓐ p 2 − 22 p p 2 − 22 p ⓑ y 2 + 5 y y 2 + 5 y ⓒ m 2 + 2 5 m m 2 + 2 5 m

ⓐ q 2 − 6 q q 2 − 6 q ⓑ x 2 − 7 x x 2 − 7 x ⓒ n 2 − 2 3 n n 2 − 2 3 n

Solve Quadratic Equations of the form x 2 + bx + c = 0 by Completing the Square

In the following exercises, solve by completing the square.

u 2 + 2 u = 3 u 2 + 2 u = 3

z 2 + 12 z = −11 z 2 + 12 z = −11

x 2 − 20 x = 21 x 2 − 20 x = 21

y 2 − 2 y = 8 y 2 − 2 y = 8

m 2 + 4 m = −44 m 2 + 4 m = −44

n 2 − 2 n = −3 n 2 − 2 n = −3

r 2 + 6 r = −11 r 2 + 6 r = −11

t 2 − 14 t = −50 t 2 − 14 t = −50

a 2 − 10 a = −5 a 2 − 10 a = −5

b 2 + 6 b = 41 b 2 + 6 b = 41

x 2 + 5 x = 2 x 2 + 5 x = 2

y 2 − 3 y = 2 y 2 − 3 y = 2

u 2 − 14 u + 12 = −1 u 2 − 14 u + 12 = −1

z 2 + 2 z − 5 = 2 z 2 + 2 z − 5 = 2

r 2 − 4 r − 3 = 9 r 2 − 4 r − 3 = 9

t 2 − 10 t − 6 = 5 t 2 − 10 t − 6 = 5

v 2 = 9 v + 2 v 2 = 9 v + 2

w 2 = 5 w − 1 w 2 = 5 w − 1

x 2 − 5 = 10 x x 2 − 5 = 10 x

y 2 − 14 = 6 y y 2 − 14 = 6 y

( x + 6 ) ( x − 2 ) = 9 ( x + 6 ) ( x − 2 ) = 9

( y + 9 ) ( y + 7 ) = 80 ( y + 9 ) ( y + 7 ) = 80

( x + 2 ) ( x + 4 ) = 3 ( x + 2 ) ( x + 4 ) = 3

( x − 2 ) ( x − 6 ) = 5 ( x − 2 ) ( x − 6 ) = 5

Solve Quadratic Equations of the form ax 2 + bx + c = 0 by Completing the Square

3 m 2 + 30 m − 27 = 6 3 m 2 + 30 m − 27 = 6

2 x 2 − 14 x + 12 = 0 2 x 2 − 14 x + 12 = 0

2 n 2 + 4 n = 26 2 n 2 + 4 n = 26

5 x 2 + 20 x = 15 5 x 2 + 20 x = 15

2 c 2 + c = 6 2 c 2 + c = 6

3 d 2 − 4 d = 15 3 d 2 − 4 d = 15

2 x 2 + 7 x − 15 = 0 2 x 2 + 7 x − 15 = 0

3 x 2 − 14 x + 8 = 0 3 x 2 − 14 x + 8 = 0

2 p 2 + 7 p = 14 2 p 2 + 7 p = 14

3 q 2 − 5 q = 9 3 q 2 − 5 q = 9

5 x 2 − 3 x = −10 5 x 2 − 3 x = −10

7 x 2 + 4 x = −3 7 x 2 + 4 x = −3

Writing Exercises

Solve the equation x 2 + 10 x = −25 x 2 + 10 x = −25

ⓐ by using the Square Root Property

ⓑ by Completing the Square

ⓒ Which method do you prefer? Why?

Solve the equation y 2 + 8 y = 48 y 2 + 8 y = 48 by completing the square and explain all your steps.

ⓐ After completing the exercises, use this checklist to evaluate your mastery of the objectives of this section.

ⓑ After reviewing this checklist, what will you do to become confident for all objectives?

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• Publisher/website: OpenStax
• Book title: Intermediate Algebra 2e
• Publication date: May 6, 2020
• Location: Houston, Texas
• Book URL: https://openstax.org/books/intermediate-algebra-2e/pages/1-introduction
• Section URL: https://openstax.org/books/intermediate-algebra-2e/pages/9-2-solve-quadratic-equations-by-completing-the-square

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Algebra I (2018 edition)

Course: algebra i (2018 edition)   >   unit 16.

• Quiz 3 Quadratics

Solving Quadratics by Completing the Square Activity - Digital Card Game

• Google Drive™ folder

Description

It's the Luck of the Draw with this Solving Quadratics by Completing the Square self-checking digital card game activity where students will be solving quadratic equations using the method of completing the square. This card game will be a splash in your classroom featuring 12 questions!

For students, Luck of the Draw makes mastering completing the square a ton of fun while ensuring students enjoy a unique competitive Card Game experience!

For you, not only will you enjoy the smiles on your students faces, but even better will be the time saved on prepping and grading with this no-prep self-grading digital activity!

How the activity works

In the game "Luck of the Draw" students will test their luck and their knowledge of completing the square. They will receive instant feedback and be dealt a new card with each correct answer, building hands of four-cards.

They will rack up points by getting sets of the same card or by getting a color-streak.

The card deck is made up of 64 cards, 4 colors, and 16 unique pool/beach themed images.

As they answer all 12 questions in this activity, students will make 3 sets of 4 cards and earn points for each hand. At the end, they will be given bonus points by combining all 3 hands into one mega-hand.

The student with the most points at the end wins. You can have students compete in small groups, pairs, or as a whole class.

Additional Features:

• Grade Score % on assignment displays at the top of the page
• Task card problems appear as drop down menus, ensuring each question is easily readable
• Correct answers will turn green
• Incorrect answers will turn red
• Each student copy will generate a unique set of cards (or use the refresh button to shuffle and play again)
• Automatically tallies up points after each round and keeps track of final score

This innovative and gamified approach to learning will captivate your students' imaginations and create a classroom atmosphere brimming with excitement and enthusiasm. Get ready for the splash this engaging digital card game will make in your classroom!

Why You'll LOVE this Product

• Instant, self-grading
• Instant feedback for student
• Engaging Gamified Activity
• Digital practice (save on copies if you'd like)
• Optional student recording sheet included
• Answer key included
• Provide students the opportunity to revise and check answers with instant feedback
• Good mix and variety of questions and levels of difficulty

Concepts and Problem Types

• Solving quadratic equations by completing the square
• All problems produce integer solutions
• Quadratics where a = 1 and b is even
• Quadratics where a = 1 and b is odd
• Quadratics where a > 1 and b is even
• Includes quadratics not written in standard form

Suggestions for Use

• Pre-Assessment
• Formative Assessment
• Group/partner work
• Class competition
• Math centers
• Small group instruction
• Extra help or practice
• Mid-unit practice
• End of unit review
• In-class practice
• Early-finisher work
• Extra credit
• Over-break practice assignment

What is Included in the Google Drive Folder

• Student version of digital game activity
• Instructions and point scoring break down (if you or students want to know how their score is calculated)
• Optional printable student recording sheet

Google Activities Tips

This digital resource was designed using Google Sheets™ and can be used on Google Classroom and Google Drive.

If you haven't created a free Google account, you will need to do that before opening this activity. Each student will need their own Google Account to open this activity. Once you have added the resource to your google drive you can assign the activity to your students. When assigning, make sure you select "make a copy for each student" if assigning on Google Classroom. If assigning using a different platform, ensure students are working with a COPY  of the activity.

Related Products

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⭐ Quadratics in Vertex Form Graphing Transformations Scavenger Hunt Activity

⭐ Solving Quadratic Equations by Factoring Activity Digital Partner Mystery Game

⭐ Writing Quadratic Functions in Vertex Form Partner Activity (Print & Digital)

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Copyright Terms:

The graphics and clipart in this resource as meant to be used for your classroom only. You may not extract the images for other uses.

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The Squadratic Formula

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