Lesson 9-4 Solving Quadratic Equations by Completing the Square
Completing the Square – a technique used to make a quadratic equation a perfect square
The technique is as follows:
1. Find one half of b, the coefficient of the x term
2. Square the result from Step 1
3. Add the result of Step 2 to x2+bx
Example 1 Complete the Square
Find the value of c that makes x2 – 8x + c a perfect square trinomial.
Complete the square.
Step 1
1
Find 2 of -8.
–8
2 = –4
Step 2
Square the result of Step 1.
(–4)2 = 16
Step 3
Add the result of Step 2 to x2 – 8x .
x2 – 8x + 16
Thus, c = 16. Notice that x2 – 8x + 16 = (x – 4)2.
Example 2 Solve an Equation by Completing the Square
Solve x2 + 4x + 5 = 0 by completing the square.
Isolate the x2- and x-terms. Then complete the square and solve.
x2 + 4x – 5 = 0
Original equation
2
x + 4x – 5 + 5 = 0 + 5
Add 5 to each side.
2
x + 4x = 5
Simplify.
4
2
x + 4x + 4 = 5 + 4
Since   2 = 4, add 4 to each side.
2
2
2
(x + 2) = 9
Factor x + 4x + 4.
x + 2 = 3
Take the square root of each side.
x + 2 – 2 = 3 – 2
Subtract 2 from each side.
x = –2  3
Simplify.
x = –2 + 3 or x = –2 – 3
Separate the solutions.
=1
= –5
Simplify.
The solutions are 1 and –5.
Example 3 Equation with a  1
Solve 5x2 – 80x – 5 = 0 by completing the square.
5x2 – 80x – 5 = 0
5 x 2 80 x  5 0
=
5
5
2
x – 16x – 1 = 0
x2 – 16x – 1 + 1 = 0 + 1
x2 – 16x = 1
x2 – 16x + 64 = 1 + 64
(x – 8)2 = 65
x – 8 =  65
x – 8 + 8 =  65 + 8
x = 8  65
Original equation
Divide each side by 5.
Simplify.
Add 1 to each side.
Simplify.
16  2 = 64, add 64 to each side.
Since 

 2 
2
Factor x – 16x + 64.
Take the square root of each side.
Add 8 to each side.
Simplify.
Use a calculator to evaluate each value of x.
x = 8 + 65 or x = 8 – 65
 16.1
 –0.1
The solutions are approximately 16.1 and –0.1.
Real-World Example 4 Solve a Problem by Completing the Square
GEOMETRY The area of the rectangle at the
right is 144 square centimeters. What are the
length and width of the rectangle?
Write an equation for the area of the rectangle. Then
complete the square to solve for x.
A=w
Formula for area of a rectangle
144 = 3x(x – 8)
Substitution
144 = 3x2 – 24x
Simplify.
2
144 3x – 24x
Divide each side by 3.
3 =
3
48 = x2 – 8x
Simplify.
(x – 8) cm
3x cm
2
48 + 16 = x2 – 8x + 16
64 = x2 – 8x + 16
64 = (x – 4)2
8 = x – 4
48=x
x = 12 or x = –4
 8 
Since   = 16, add 16 to each side.
 2 
Simplify.
2
Factor x – 8x + 16.
Take the square root of each side.
Add 4 to each side.
Simplify.
Since a rectangle cannot have negative dimensions, the negative solution is not reasonable.
So, the rectangle is 3(12) or 36 centimeters long and 12 – 8 or 4 centimeters wide.