Warm Up
Simplify.
1.
2.
3.
4.
19
Warm Up
Solve each quadratic equation by
factoring.
5. x2 + 8x + 16 = 0 x = –4
6. x2 – 22x + 121 = 0 x = 11
7. x2 – 12x + 36 = 0
x=6
Objective
Solve quadratic equations by
completing the square.
Vocabulary
completing the 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 + 9
x2 – 8x + 16
Divide the coefficient of
the x-term by 2, then
square the result to get
the constant term.
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.
Example 1: Completing the Square
Complete the square to form a perfect square
trinomial.
A. x2 + 2x +
x2 + 2x
B. x2 – 6x +
Identify b.
x2 + –6x
.
x2 + 2x + 1
x2 – 6x + 9
Example 1
Complete the square to form a perfect square
trinomial.
a. x2 + 12x +
x2 + 12x
b. x2 – 5x +
Identify b.
x2 + –5x
.
x2 + 12x + 36
x2 – 6x +
Example 1
Complete the square to form a perfect square
trinomial.
c. 8x + x2 +
x2 + 8x
Identify b.
.
x2 + 12x + 16
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.
Solving a Quadratic Equation by Completing the Square
Example 2A: Solving x2 +bx = c
Solve by completing the square.
x2 + 16x = –15
The equation is in the
Step 1 x2 + 16x = –15
form x2 + bx = c.
Step 2
.
Step 3 x2 + 16x + 64 = –15 + 64
Complete the square.
Step 4 (x + 8)2 = 49
Factor and simplify.
Step 5 x + 8 = ± 7
Take the square root
of both sides.
Write and solve two
equations.
Step 6 x + 8 = 7 or x + 8 = –7
x = –1 or
x = –15
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)
1 – 16
–15
–15
–15
–15
x2 + 16x = –15
(–15)2 + 16(–15)
225 – 240
–15
–15
–15
–15
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.
Step 5 x – 2 = ± √10
Take the square root
of both sides.
Step 6 x – 2 = √10 or x – 2 = –√10 Write and solve two
x = 2 + √10 or x = 2 – √10 equations.
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.
Check It Out! Example 2a
Solve by completing the square.
x2 + 10x = –9
Step 1 x2 + 10x = –9
Step 2
Step 3 x2 + 10x + 25 = –9 + 25
Step 4 (x + 5)2 = 16
Step 5 x + 5 = ± 4
Step 6 x + 5 = 4 or x + 5 = –4
x = –1 or
x = –9
The equation is in the
form x2 + bx = c.
.
Complete the square.
Factor and simplify.
Take the square root
of both sides.
Write and solve two
equations.
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)
1 – 16
–15
–15
–15
–15
x2 + 10x = –9
(–9)2 + 10(–9)
–9
81 – 90
–9
–9
–9

Check It Out! Example 2b
Solve by completing the square.
t2 – 8t – 5 = 0
Step 1 t2 + (–8t) = 5
Step 2
Write in the form
x2 + bx = c.
.
Step 3 t2 – 8t + 16 = 5 + 16
Complete the square.
Step 4 (t – 4)2 = 21
Factor and simplify.
Step 5 t – 4 = ± √21
Take the square root
of both sides.
Step 6 t = 4 + √21 or t = 4 – √21
Write and solve two
equations.
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.
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
x2 + (–4x) = –5
Step 2
Write in the form x2 + bx = c.
.
Step 3 x2 – 4x + 4 = –5 + 4 Complete the square.
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.
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
.
Example 3B Continued
Solve by completing the square.
Step 3
Complete the square.
Rewrite using like
denominators.
Step 4
Factor and simplify.
Step 5
Take the square root
of both sides.
Example 3B Continued
Solve by completing the square.
Write and solve
two equations.
Step 6
The solutions are
and –4.
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.
Example 3a Continued
Solve by completing the square.
Step 2
.
Step 3
Complete the square.
Step 4
Factor and simplify.
Example 3a Continued
Solve by completing the square.
Step 5
Take the square root
of both sides.
Step 6
Write and solve two
equations.
−
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.
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.
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.
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.
Example 4 Continued
3
Solve
Let x be the width.
Then x + 2 is the length.
Use the formula for area of a rectangle.
•
w
=
A
l
length
x+2
times
•
width
x
= area of room
=
195
Example 4 Continued
Step 1 x2 + 2x = 195
Step 2
Simplify.
.
Step 3 x2 + 2x + 1 = 195 + 1 Complete the square by
adding 1 to both sides.
Step 4 (x + 1)2 = 196
Factor the perfect-square
trinomial.
Take the square root of
Step 5 x + 1 = ± 14
both sides.
Step 6 x + 1 = 14 or x + 1 = –14 Write and solve two
equations.
x = 13 or x = –15
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 13 + 2, or
15, feet.
4
Look Back
The length of the room is 2 feet greater than the
width. Also 13(15) = 195.
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.
Example 4 Continued
2
Make a Plan
Set the formula for the area of a rectangle equal
to 400, the area of the room.
Solve the equation.
Example 4 Continued
3
Solve
Let x be the width.
Then x + 8 is the length.
Use the formula for area of a rectangle.
l
length
X+8
•
times
•
w
=
width
= area of room
x
=
A
400
Example 4 Continued
Step 1 x2 + 8x = 400
Step 2
Simplify.
.
Step 3 x2 + 8x + 16 = 400 + 16 Complete the square by
adding 16 to both sides.
Step 4 (x + 4)2 = 416
Factor the perfectsquare 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
x ≈ 16.4 or x ≈ –24.4
equations.
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 16.4 + 8, 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) = 400.16, which is
approximately 400.
Lesson Quiz: Part I
Complete the square to form a perfect square
trinomial.
1. x2 +11x +
2. x2 – 18x +
81
Solve by completing the square.
3. x2 – 2x – 1 = 0
4. 3x2 + 6x = 144
5. 4x2 + 44x = 23
6, –8
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