1. Warm Up
Simplify. 1. 2. 19 3. 4.

2. Warm Up
Solve each quadratic equation by factoring.
5. x2 + 8x + 16 = 0
6. x2 – 22x = 0
7. x2 – 12x + 36 = 0
x = –4
x = 11
x = 6

3. Objective
Solve quadratic equations by completing the square.

4. Vocabulary
completing the square

5. In the previous lesson, you solved quadratic equations by isolating x2 and then using square roots. This method works if the quadratic equation, when written in standard form, is a perfect square.
When a trinomial is a perfect square, there is a relationship between the coefficient of the x-term and the constant term.
X2 + 6x x2 – 8x + 16
Divide the coefficient of the x-term by 2, then square the result to get the constant term.

6. An expression in the form x2 + bx is not a perfect square
An expression in the form x2 + bx is not a perfect square. However, you can use the relationship shown above to add a term to x2 + bx to form a trinomial that is a perfect square. This is called completing the square.

7. Example 1: Completing the Square
Complete the square to form a perfect square trinomial.
A. x2 + 2x +
B. x2 – 6x +
x2 + 2x
x2 + –6x
Identify b. .
x2 + 2x + 1
x2 – 6x + 9

8. Check It Out! Example 1
Complete the square to form a perfect square trinomial.
a. x2 + 12x +
b. x2 – 5x +
x2 + 12x
x2 + –5x
Identify b. .
x2 – 6x +
x2 + 12x + 36

9. Check It Out! Example 1
Complete the square to form a perfect square trinomial.
c. 8x + x2 +
x2 + 8x
Identify b. .
x2 + 12x + 16

10. To solve a quadratic equation in the form x2 + bx = c, first complete the square of x2 + bx. Then you can solve using square roots.

11. Solving a Quadratic Equation by Completing the Square

12. Example 2A: Solving x2 +bx = c
Solve by completing the square.
x2 + 16x = –15
The equation is in the form x2 + bx = c.
Step 1 x2 + 16x = –15
Step 2 .
Step 3 x2 + 16x + 64 = –
Complete the square.
Step 4 (x + 8)2 = 49
Factor and simplify.
Take the square root of both sides.
Step 5 x + 8 = ± 7
Step 6 x + 8 = 7 or x + 8 = –7
x = –1 or x = –15
Write and solve two equations.

13.   Example 2A Continued Solve by completing the square.
x2 + 16x = –15
The solutions are –1 and –15.
Check
x2 + 16x = –15
(–1)2 + 16(–1) –15
1 – –15
– –15 
x2 + 16x = –15
(–15)2 + 16(–15) –15
225 – –15
– –15 

14. Example 2B: Solving x2 +bx = c
Solve by completing the square.
x2 – 4x – 6 = 0
Write in the form x2 + bx = c.
Step 1 x2 + (–4x) = 6
Step 2 .
Step 3 x2 – 4x + 4 = 6 + 4
Complete the square.
Step 4 (x – 2)2 = 10
Factor and simplify.
Take the square root of both sides.
Step 5 x – 2 = ± √10
Step 6 x – 2 = √10 or x – 2 = –√10
x = 2 + √10 or x = 2 – √10
Write and solve two equations.

15. Example 2B Continued
Solve by completing the square.
The solutions are 2 + √10 and x = 2 – √10.
Check Use a graphing calculator to check your answer.

16. Check It Out! Example 2a
Solve by completing the square.
x2 + 10x = –9
The equation is in the form x2 + bx = c.
Step 1 x2 + 10x = –9
Step 2 .
Step 3 x2 + 10x + 25 = –9 + 25
Complete the square.
Factor and simplify.
Step 4 (x + 5)2 = 16
Take the square root of both sides.
Step 5 x + 5 = ± 4
Step 6 x + 5 = 4 or x + 5 = –4
x = –1 or x = –9
Write and solve two equations.

17. Check It Out! Example 2a Continued
Solve by completing the square.
x2 + 10x = –9
The solutions are –9 and –1.
Check
x2 + 16x = –15
(–1)2 + 16(–1) –15
1 – –15
– –15 
x2 + 10x = –9
(–9)2 + 10(–9) –9
81 – –9
–9 –9 

18. Check It Out! Example 2b
Solve by completing the square.
t2 – 8t – 5 = 0
Write in the form
x2 + bx = c.
Step 1 t2 + (–8t) = 5
Step 2 .
Step 3 t2 – 8t + 16 =
Complete the square.
Factor and simplify.
Step 4 (t – 4)2 = 21
Take the square root of both sides.
Step 5 t – 4 = ± √21
Step 6 t = 4 + √21 or t = 4 – √21
Write and solve two equations.

19. Check It Out! Example 2b Continued
Solve by completing the square.
The solutions are
t = 4 – √21 or t = 4 + √21.
Check Use a graphing calculator to check your answer.

20. Example 3A: Solving ax2 + bx = c by Completing the Square
Solve by completing the square.
–3x2 + 12x – 15 = 0
Step 1
Divide by – 3 to make a = 1.
x2 – 4x + 5 = 0
x2 – 4x = –5
Write in the form x2 + bx = c.
x2 + (–4x) = –5
Step 2 .
Step 3
x2 – 4x + 4 = –5 + 4
Complete the square.

21. Example 3A Continued
Solve by completing the square.
–3x2 + 12x – 15 = 0
Step 4
(x – 2)2 = –1
Factor and simplify.
There is no real number whose square is negative, so there are no real solutions.

22. Example 3B: Solving ax2 + bx = c by Completing the Square
Solve by completing the square.
5x2 + 19x = 4
Step 1
Divide by 5 to make a = 1.
Write in the form x2 + bx = c. .
Step 2

23. Example 3B Continued
Solve by completing the square.
Step 3
Complete the square.
Rewrite using like denominators.
Step 4
Factor and simplify.
Take the square root of both sides.
Step 5

24. Example 3B Continued
Solve by completing the square.
Write and solve two equations.
Step 6
The solutions are and –4.

25. Check It Out! Example 3a
Solve by completing the square.
3x2 – 5x – 2 = 0
Step 1
Divide by 3 to make a = 1.
Write in the form x2 + bx = c.

26. Check It Out! Example 3a Continued
Solve by completing the square. .
Step 2
Step 3
Complete the square.
Step 4
Factor and simplify.

27. Check It Out! Example 3a Continued
Solve by completing the square.
Step 5
Take the square root of both sides.
Step 6
Write and solve two equations. −

28. Check It Out! Example 3b
Solve by completing the square.
4t2 – 4t + 9 = 0
Step 1
Divide by 4 to make a = 1.
Write in the form x2 + bx = c.

29. Check It Out! Example 3b Continued
Solve by completing the square.
4t2 – 4t + 9 = 0 .
Step 2
Step 3
Complete the square.
Step 4
Factor and simplify.
There is no real number whose square is negative, so there are no real solutions.

30. Understand the Problem
Example 4: Problem-Solving Application
A rectangular room has an area of 195 square feet. Its width is 2 feet shorter than its length. Find the dimensions of the room. Round to the nearest hundredth of a foot, if necessary. 1
Understand the Problem
The answer will be the length and width of the room.
List the important information:
The room area is 195 square feet.
The width is 2 feet less than the length.

31. Example 4 Continued 2
Make a Plan
Set the formula for the area of a rectangle equal to 195, the area of the room.
Solve the equation.

32. Use the formula for area of a rectangle.
Example 4 Continued
Solve 3
Let x be the width.
Then x + 2 is the length.
Use the formula for area of a rectangle.
l • w = A
length
times
width =
area of room
x + 2 • x =
195

33. Example 4 Continued
Step 1 x2 + 2x = 195
Simplify. .
Step 2
Complete the square by adding 1 to both sides.
Step 3 x2 + 2x + 1 =
Step 4 (x + 1)2 = 196
Factor the perfect-square trinomial.
Take the square root of both sides.
Step 5 x + 1 = ± 14
Step 6 x + 1 = 14 or x + 1 = –14
Write and solve two equations.
x = 13 or x = –15

34. Example 4 Continued
Negative numbers are not reasonable for length, so x = 13 is the only solution that makes sense.
The width is 13 feet, and the length is , or 15, feet. 4
Look Back
The length of the room is 2 feet greater than the width. Also 13(15) = 195.

35. Understand the Problem
Check It Out! Example 4
An architect designs a rectangular room with an area of 400 ft2. The length is to be 8 ft longer than the width. Find the dimensions of the room. Round your answers to the nearest tenth of a foot. 1
Understand the Problem
The answer will be the length and width of the room.
List the important information:
The room area is 400 square feet.
The length is 8 feet more than the width.

36. Check It Out! Example 4 Continued2
Make a Plan
Set the formula for the area of a rectangle equal to 400, the area of the room.
Solve the equation.

37. Check It Out! Example 4 Continued
Solve 3
Let x be the width.
Then x + 8 is the length.
Use the formula for area of a rectangle.
l • w = A
length
times
width =
area of room
X + 8 • x =
400

38. Check It Out! Example 4 Continued
Step 1 x2 + 8x = 400
Simplify.
Step 2 .
Step 3 x2 + 8x + 16 =
Complete the square by adding 16 to both sides.
Step 4 (x + 4)2 = 416
Factor the perfect-square trinomial.
Step 5 x + 4  ± 20.4
Take the square root of both sides.
Step 6 x + 4  20.4 or x + 4  –20.4
Write and solve two equations.
x  16.4 or x  –24.4

39. Check It Out! Example 4 Continued
Negative numbers are not reasonable for length, so x  16.4 is the only solution that makes sense.
The width is approximately16.4 feet, and the length is , or approximately 24.4, feet. 4
Look Back
The length of the room is 8 feet longer than the width. Also 16.4(24.4) = , which is approximately 400.

40. Lesson Quiz: Part I
Complete the square to form a perfect square trinomial.
1. x2 +11x +
2. x2 – 18x +
Solve by completing the square.
3. x2 – 2x – 1 = 0
4. 3x2 + 6x = 144
5. 4x x = 23 81
6, –8

41. Lesson Quiz: Part II
6. Dymond is painting a rectangular banner for a football game. She has enough paint to cover 120 ft2. She wants the length of the banner to be 7 ft longer than the width. What dimensions should Dymond use for the banner?
8 feet by 15 feet