Completing the Square

9.2.3 Completing the Square

9.2.3 • Completing the Square

Activity

One technique for solving quadratic equations is called completing the square. Here are two examples of how Diego and Mai completed the square to solve the same equation.

Diego	Mai
$\begin{array}{rcl} x^{2} + 10 x + 9 & = & 0 \\ x^{2} + 10 x & = & - 9 \\ x^{2} + 10 x + 25 & = & - 9 + 25 \\ x^{2} + 10 x + 25 & = & 16 \\ (x + 5)^{2} & = & 16 \\ \sqrt{(x + 5)^{2}} & = & \sqrt{16} \\ x + 5 & = & \pm 4 \\ x + 5 = 4 & or & x + 5 = - 4 \\ x = - 1 & and & x = - 9 \end{array}$	$\begin{array}{rcl} x^{2} + 10 x + 9 & = & 0 \\ x^{2} + 10 x + 9 + 16 & = & 16 \\ x^{2} + 10 x + 25 & = & 16 \\ (x + 5)^{2} & = & 16 \\ \sqrt{(x + 5)^{2}} & = & \sqrt{16} \\ (x + 5) & = & \pm 4 \\ x + 5 = 4 & and & x + 5 = - 4 \\ x = - 1 & or & x = - 9 \end{array}$

Study the worked examples. Then, try solving these equations by completing the square for questions 1 – 5.

$x^{2} + 6 x + 8 = 0$

Compare your answer:

$x^{2} + 6 x + 8 = 0 x^{2} + 6 x = - 8 x^{2} + 6 x + 9 = - 8 + 9 (x + 3)^{2} = 1 \sqrt{(x + 3)^{2}} = \sqrt{1} x + 3 = \pm 1 x + 3 = 1 x + 3 = - 1 x = - 4 and x = - 2$

$x^{2} + 12 x = 13$

Compare your answer:

$x^{2} + 12 x = 13 x^{2} + 12 x + 36 = 13 + 36 x^{2} + 12 x + 36 = 49 (x + 6)^{2} = 49 \sqrt{(x + 6)^{2}} = \sqrt{49} x + 6 = \pm 7 x + 6 = 7 x + 6 = - 7 x = - 13 and x = 1$

$0 = x^{2} - 10 x + 21$

Compare your answer:

$0 = x^{2} - 10 x + 21 x^{2} - 10 x + 21 = 0 x^{2} - 10 x = - 21 x^{2} - 10 x + 25 = - 21 + 25 x^{2} - 10 x + 25 = 4 (x - 5)^{2} = 4 \sqrt{(x - 5)^{2}} = \sqrt{4} x - 5 = \pm 2 x - 5 = 2 x - 5 = - 2 x = 3 and x = 7$

$x^{2} - 2 x + 3 = 83$

Compare your answer:

$x^{2} - 2 x + 3 = 83 x^{2} - 2 x = 83 - 3 x^{2} - 2 x = 80 x^{2} - 2 x + 1 = 80 + 1 x^{2} - 2 x + 1 = 81 \sqrt{(x - 1)^{2}} = \sqrt{81} x - 1 = \pm 9 x - 1 = 9 x - 1 = - 9 x = - 8 and x = 10$

$x^{2} + 40 = 14 x$

Compare your answer:

$x^{2} + 40 = 14 x x^{2} - 14 x + 40 = 0 x^{2} - 14 x = - 40 x^{2} - 14 x + 49 = - 40 + 49 x^{2} - 14 x + 49 = 9 \sqrt{(x - 7)^{2}} = \sqrt{9} x - 7 = \pm 3 x - 7 = 3 x - 7 = - 3 x = 4 and x = 10$

Video: Solve Equations by Completing the Square

Watch the following video to learn more about solving equations by completing the square.

Access multimedia content

Are you ready for more?

Extending Your Thinking

Here is a diagram made of a square and two congruent rectangles. Its total area is $x^{2} + 35 x$ square units.

Diagram A

A diagram consisting of a square of side length x and two rectangles. The square is placed in the upper left corner of a figure with a rectangle extending from one side of the square vertically down and the other extending horizontally to the right. The three figures together create a large right angle where the rectangles represent the sides of the right angle and the square represents the vertex.

1. What is the length of the unlabeled side of each of the two rectangles in Diagram A?

Compare your answer:

17.5

2. In Diagram B, we add lines to make the figure a square. The area of the square becomes a quadratic equation.

A figure consists of a small square of side length x and two congruent rectangles that extend horizontally and vertically from its sides. In addition, a larger square fits in the area between the congruent rectangles.

Calculate the equation, then put that equation in the standard form $a x^{2} + b x + c$ to answer questions a and b.

a. Using the standard form $a x^{2} + b x + c$ to represent the area of the square in Diagram B, what is the value of the coefficient $b$ ?

Compare your answer:

b. Using the standard form $a x^{2} + b x + c$ to represent the area of the square in Diagram B, what is the value of the constant $c$ ?

Compare your answer:

306.25

3. How is the process of finding the area of the entire figure like the process of building perfect squares for expressions like $x^{2} + b x$ ?

Compare your answer:

First, we had to figure out what number was half of 35 (which is 17.5) to find the side length of the rectangles. Then, we had to square the number 17.5 to find the area of the missing piece.

Self Check

Find the solutions to

x^2-10x=-9

by completing the square.

$x=9$
$x=21$
$x=1$ , $x=9$
$x=25$

Additional Resources

Solving Quadratic Equations by Completing the Square

Example

Solve $x^{2} + 8 x = 48$ by completing the square.

Step 1 - Isolate the variable terms on one side and place the constant term on the other. This equation has all the variables on the left.

$a x^{2} + b x = c$

$x^{2} + 8 x = 48$

Step 2 - Find ( $\frac{b}{2})^{2}$ , the number to complete the square. Add to both sides of the equation. Take half of 8 and square it.

$4^{2} = 16$

Add 16 to BOTH sides of the equation.

$x^{2} + 8 x +____= 48$

$(\frac{b}{2})^{2}$

$x^{2} + 8 x + 16 = 48 + 16$

Step 3 - Factor the perfect square trinomial as a binomial square.

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

Substitute the binomial square term for the trinomial.

$(x + 4)^{2} = 64$

Step 4 - Use the square root property.

$\sqrt{(x + 4)^{2}} = \sqrt{64}$

$x + 4 = \pm 8$

Step 5 - Simplify the radical and then solve the two resulting equations.

$x + 4 = \pm 8$ $x + 4 = 8$

$x + 4 = - 8$ $x = 4$

$x = - 12$

Step 6 - Check the solutions. Put each answer in the original equation to check.

Substitute $x = 4$ .

$x^{2} + 8 x = 48$

$(4)^{2} + 8 (4) ≟ 48$

$16 + 32 ≟ 48$

$48 = 48 ✔$

Substitute $x = - 12$ .

$x^{2} + 8 x = 48$

$(- 12)^{2} + 8 (- 12) ≟ 48$

$144 - 96 ≟ 48$

$48 = 48 ✔$

Try it

Solving Quadratic Equations by Completing the Square

Solve $x^{2} + 4 x = 5$ by completing the square.

Here is how to solve this quadratic using completing the square:

Step 1 - Isolate the constants on one side of the equation.

$x^{2} + 4 x = 5$

Step 2 - Find $(\frac{b}{2})^{2}$ . Add this number to both sides of the equation.

$b = 4$ ,

$(\frac{4}{2})^{2} = 2^{2} = 4$

$x^{2} + 4 x + 4 = 5 + 4$

$x^{2} + 4 x + 4 = 9$

Step 3 - Factor the perfect square trinomial.

$(x + 2)^{2} = 9$

Step 4 - Use the square root property.

$\sqrt{(x + 2)^{2}} = 9$

$x + 2 = \pm 3$

Step 5 - Separate into two equations and solve.

$x + 2 = - 3$

$x + 2 = 3$

$x = - 5$

$x = 1$

Step 6 - Check into the original equation.

$x = - 5$

$(- 5)^{2} + 4 (- 5) = 5$

$25 + - 20 = 5$

$x = 1$

$1^{2} + 4 (1) = 5$

$5 = 5$