1. Complete the Square Lesson 1.7
Honors Algebra 2
Complete the Square
Lesson 1.7

2. Goals Goal Rubric Complete the Square.
Solve Quadratic Equations by completing the square.
Changing Quadratic Equations to Vertex Form by Completing the Square.
Level 1 – Know the goals.
Level 2 – Fully understand the goals.
Level 3 – Use the goals to solve simple problems.
Level 4 – Use the goals to solve more advanced problems.
Level 5 – Adapts and applies the goals to different and more complex problems.

3. Vocabulary
Completing the square

4. Solve Quadratic Equation by Finding Square Roots
Previously, solved equations of the form 𝑥 2 =𝑘 by finding square roots. Finding square roots also works if one side of an equation is a perfect square trinomial, 𝑎 2 +2𝑎𝑏+ 𝑏 2 or 𝑎 2 −2𝑎𝑏+ 𝑏 2 .

5. The solutions are 4 + 5 = 9 and 4 –5 = – 1.
EXAMPLE 1
Solve a quadratic equation by finding square roots
Solve x2 – 8x + 16 = 25.
x2 – 8x + 16 = 25
Write original equation.
(x – 4)2 = 25
Write left side as a binomial squared.
x – 4 = + 5
Take square roots of each side.
x = 4 + 5
Solve for x.
The solutions are = 9 and 4 –5 = – 1.
ANSWER

6. Solve the equation by finding square roots.
Your Turn:
for Example 1
Solve the equation by finding square roots.
x2 + 6x + 9 = 36
The solutions are 3 and – 9.
ANSWER
x2 – 10x + 25 = 1
The solutions are 4 and 6.
ANSWER
x2 – 24x = 100
The solutions are 2 and 22.
ANSWER

7. Completing the Square
The methods in the previous examples can be used only for expressions that are perfect squares. However, you can use algebra to rewrite any quadratic expression as a perfect square.
You can use algebra tiles to model a perfect square trinomial as a perfect square. The area of the square at right is x2 + 2x + 1. Because each side of the square measures x + 1 units, the area is also (x + 1)(x + 1), or (x + 1)2. This shows that (x + 1)2 = x2 + 2x + 1.

8. Completing the Square *** Important To Remember ***
If a quadratic expression of the form x2 + bx cannot model a square, you can add a term to form a perfect square trinomial. This is called completing the square.
*** Important To Remember ***
To complete the square add 𝑏 2 2

9. Completing the Square
The model shows completing the square for x2 + 6x by adding 9 unit tiles. The resulting perfect square trinomial is x2 + 6x + 9. Note that completing the square does not produce an equivalent expression.

10. Find half the coefficient of x.
EXAMPLE 2
Make a perfect square trinomial
Find the value of c that makes x2 + 16x + c a perfect square trinomial. Then write the expression as the square of a binomial.
SOLUTION
STEP 1 16 2 = 8
Find half the coefficient of x.
STEP 2
Square the result of Step 1.
82 = 64
STEP 3
Replace c with the result of Step 2.
x2 + 16x + 64

11. EXAMPLE 2
Make a perfect square trinomial
ANSWER
The trinomial x2 + 16x + c is a perfect square when c = 64. Then x2 + 16x + 64 = (x + 8)(x + 8) = (x + 8)2.

12. The trinomial x2 + 14x + c is a perfect square when c = 49.
Your Turn:
for Example 2
Find the value of c that makes the expression a perfect square trinomial. Then write the expression as the square of a binomial. 4.
x2 + 14x + c
ANSWER
The trinomial x2 + 14x + c is a perfect square when c = 49. 5.
x2 + 22x + c
ANSWER
The trinomial x2 + 22x + c is a perfect square when c = 144. 6.
x2 – 9x + c
ANSWER
The trinomial x2 – 9x + c is a perfect square when c = 81 4

13. Solving Equations by Completing the Square
The method of completing the square can be used to solve any quadratic equation.

14. ( ) Solve x2 – 12x + 4 = 0 by completing the square. x2 – 12x + 4 = 0
EXAMPLE 3
Solve ax2 + bx + c = 0 when a = 1
Solve x2 – 12x + 4 = 0 by completing the square.
x2 – 12x + 4 = 0
Write original equation.
x2 – 12x = – 4
Write left side in the form x2 + bx.
Add
–12 2 ( ) =
(–6) 36
to each side.
x2 – 12x + 36 = –
(x – 6)2 = 32
Write left side as a binomial squared.
x – 6 =
Take square roots of each side.
x =
Solve for x.
x =
Simplify: 32 = 16 2 4
The solutions are 6 + 4
and 6 – 4 2
ANSWER .

15. You can use algebra or a graph.
EXAMPLE 3
Solve ax2 + bx + c = 0 when a = 1
CHECK
You can use algebra or a graph.
Algebra Substitute each solution in the original equation to verify that it is correct.
Graph Use a graphing calculator to graph y = x2 – 12x The x-intercepts are about – and

16. ( ) Solve ax2 + bx + c = 0 when a ≠ 1 2x2 + 8x + 14 = 0
EXAMPLE 4
Solve ax2 + bx + c = 0 when a ≠ 1
Solve 2x2 + 8x + 14 = 0 by completing the square.
2x2 + 8x + 14 = 0
Write original equation.
x2 + 4x + 7 = 0
Divide each side by the coefficient of x2.
x2 + 4x = – 7
Write left side in the form x2 + bx.
Add 4 2 ( ) =
to each side.
x2 – 4x + 4 = – 7 + 4
(x + 2)2 = –3
Write left side as a binomial squared.
x + 2 = –3
Take square roots of each side.
x = – –3
Solve for x.
x = –2 + i 3
Write in terms of the imaginary unit i.

17. Solve ax2 + bx + c = 0 when a ≠ 1 The solutions are –2 + i 3
EXAMPLE 4
Solve ax2 + bx + c = 0 when a ≠ 1
The solutions are –2 + i 3
and – 2 – i
3 .
ANSWER

18. Use the formula for the area of a rectangle to write an equation.
EXAMPLE 5
Standardized Test Practice
SOLUTION
Use the formula for the area of a rectangle to write an equation.

19. ( ) 3x(x + 2) = 72 3x2 + 6x = 72 x2 + 2x = 24 x2 + 2x + 1 = 24 + 1
EXAMPLE 5
Standardized Test Practice
3x(x + 2) = 72
Length Width = Area
3x2 + 6x = 72
Distributive property
x2 + 2x = 24
Divide each side by the coefficient of x2.
Add 2 ( ) = 1
to each side.
x2 + 2x + 1 =
(x + 1)2 = 25
Write left side as a binomial squared.
x + 1 = + 5
Take square roots of each side.
x = –1 + 5
Solve for x.

20. The value of x is 4. The correct answer is B.
EXAMPLE 5
Standardized Test Practice
So, x = –1 + 5 = 4 or x = – 1 – 5 = – 6. You can reject x = – 6 because the side lengths would be – 18 and – 4, and side lengths cannot be negative.
The value of x is 4. The correct answer is B.
ANSWER

21. Solve x2 + 6x + 4 = 0 by completing the square.
Your Turn:
for Examples 3, 4 and 5 7.
Solve x2 + 6x + 4 = 0 by completing the square.
The solutions are – 3+
and – 3 – 5
ANSWER 8.
Solve x2 – 10x + 8 = 0 by completing the square.
The solutions are 5 +
and 5 – 17
ANSWER 9.
Solve 2n2 – 4n – 14 = 0 by completing the square.
The solutions are 1 + 2
and 1 –2 2
ANSWER

22. Solve the equation by completing the square.
Your Turn:
for Examples 3, 4 and 5
Solve the equation by completing the square.
10.
3x2 + 12x – 18 = 0.
ANSWER
The solutions are – 2 + 10
and –2 –
11.
6x(x + 8) = 12
ANSWER
The solutions are – 4 +3 2
and – 4 – 3 2
12.
4p(p – 2) = 100
ANSWER
The solutions are 1 + 26
and 1 – 26

23. Changing to Vertex Form by Completing the Square
Recall the vertex form of a quadratic function, f(x) = a(x – h)2 + k, where the vertex is (h, k). Vertex form contains a perfect square, 𝑥−ℎ 2 . Therefore, you can complete the square to rewrite any quadratic function in vertex form.

24. ( ) Write y = x2 – 10x + 22 in vertex form. Then identify the vertex.
EXAMPLE 6
Write a quadratic function in vertex form
Write y = x2 – 10x + 22 in vertex form. Then identify the vertex.
y = x2 – 10x + 22
Write original function.
y + ? = (x2 –10x + ? ) + 22
Prepare to complete the square.
Add
–10 2 ( ) =
(–5) 25
to each side.
y + 25 = (x2 – 10x + 25) + 22
y + 25 = (x – 5)2 + 22
Write x2 – 10x + 25 as a binomial squared.
y = (x – 5)2 – 3
Solve for y.
The vertex form of the function is y = (x – 5)2 – 3. The vertex is (5, – 3).
ANSWER

25. Find the maximum height of the baseball.
EXAMPLE 7
Find the maximum value of a quadratic function
The height y (in feet) of a baseball t seconds after it is hit is given by this function:
Baseball
y = –16t2 + 96t + 3
Find the maximum height of the baseball.
SOLUTION
The maximum height of the baseball is the y-coordinate of the vertex of the parabola with the given equation.

26. EXAMPLE 7
Find the maximum value of a quadratic function
y = – 16t2 + 96t +3
Write original function.
y = – 16(t2 – 6t) +3
Factor –16 from first two terms.
y +(–16)(?) = –16(t2 –6t + ? ) + 3
Prepare to complete the square.
y +(–16)(9) = –16(t2 –6t + 9 ) + 3
Add
to each side.
(–16)(9)
y – 144 = –16(t – 3)2 + 3
Write t2 – 6t + 9 as a binomial squared.
y = –16(t – 3)
Solve for y.
The vertex is (3, 147), so the maximum height of the baseball is 147 feet.
ANSWER

27. Write y = x2 – 8x + 17 in vertex form. Then identify the vertex.
Your Turn:
for Examples 6 and 7
13.
Write y = x2 – 8x + 17 in vertex form. Then identify the vertex.
The vertex form of the function is y = (x – 4) The vertex is (4, 1).
ANSWER
14.
Write y = x2 + 6x + 3 in vertex form. Then identify the vertex.
The vertex form of the function is y = (x + 3)2 – 6. The vertex is (– 3, – 6).
ANSWER

28. Write f(x) = x2 – 4x – 4 in vertex form. Then identify the vertex.
Your Turn:
for Examples 6 and 7
15.
Write f(x) = x2 – 4x – 4 in vertex form. Then identify the vertex.
The vertex form of the function is f(x) = (x – 2)2 – 8. The vertex is (2 , – 8).
ANSWER
16.
WHAT IF? In Example 7, suppose the height of the baseball is given by y = – 16t2 + 80t + 2. Find the maximum height of the baseball.
The maximum height of the baseball is 102 feet.
ANSWER