holt algebra 1 9-8 completing the square warm up simplify. 19 1.2. 3.4

Holt Algebra 1

9-8 Completing the Square

Warm Up

Simplify.

19 1. 2.

3. 4.

Holt Algebra 1


Warm Up

Solve each quadratic equation by factoring.

5. x2 + 8x + 16 = 0

6. x2 – 22x + 121 = 0

7. x2 – 12x + 36 = 0

x = –4

x = 11

x = 6

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Solve quadratic equations by completing the square.

Objective

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completing the square

Vocabulary

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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 + 9 x2 – 8x + 16 Divide the coefficient of the x-term by 2, then square the result to get the constant term.

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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.

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Example 1: Completing the Square

Complete the square to form a perfect square trinomial.

A. x2 + 2x + B. x2 – 6x +

x2 + 2x

x2 + 2x + 1

x2 + –6x

x2 – 6x + 9

Identify b.

.

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Check It Out! Example 1


a. x2 + 12x + b. x2 – 5x +

x2 + 12x

x2 + 12x + 36

x2 + –5xIdentify b.

x2 – 6x +

.

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c. 8x + x2 +

x2 + 8x

x2 + 12x + 16

Identify b.

.

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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.

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Solving a Quadratic Equation by Completing the Square

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Example 2A: Solving x2 +bx = c

Solve by completing the square.

x2 + 16x = –15

Step 1 x2 + 16x = –15

Step 2

Step 3 x2 + 16x + 64 = –15 + 64

Step 4 (x + 8)2 = 49

Step 5 x + 8 = ± 7

Step 6 x + 8 = 7 or x + 8 = –7

x = –1 or x = –15

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.

.

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Example 2A Continued


x2 + 16x = –15

The solutions are –1 and –15.

Check x2 + 16x = –15

(–1)2 + 16(–1) –15

1 – 16 –15

–15 –15

x2 + 16x = –15

(–15)2 + 16(–15) –15

225 – 240 –15

–15 –15

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Example 2B: Solving x2 +bx = c


x2 – 4x – 6 = 0

Step 1 x2 + (–4x) = 6

Step 3 x2 – 4x + 4 = 6 + 4

Step 4 (x – 2)2 = 10

Step 5 x – 2 = ± √10

Write in the form x2 + bx = c.





Step 6 x – 2 = √10 or x – 2 = –√10

x = 2 + √10 or x = 2 – √10

.Step 2

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Example 2B Continued


The solutions are 2 + √10 and x = 2 – √10.

Check Use a graphing calculator to check your answer.

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Check It Out! Example 2a


x2 + 10x = –9

Step 1 x2 + 10x = –9

Step 3 x2 + 10x + 25 = –9 + 25 Complete the square.

The equation is in the form x2 + bx = c.

Step 2

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




.

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Check It Out! Example 2a Continued


x2 + 10x = –9

The solutions are –9 and –1.

x2 + 10x = –9

(–9)2 + 10(–9) –9

81 – 90 –9

–9 –9

x2 + 16x = –15

(–1)2 + 16(–1) –15

1 – 16 –15

–15 –15

Check

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Check It Out! Example 2b


t2 – 8t – 5 = 0

Step 1 t2 + (–8t) = 5

Step 3 t2 – 8t + 16 = 5 + 16 Complete the square.


Step 2

Step 4 (t – 4)2 = 21

Step 5 t – 4 = ± √21




Step 6 t = 4 + √21 or t = 4 – √21

.

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Check It Out! Example 2b Continued


t = 4 – √21 or t = 4 + √21.The solutions are

Check Use a graphing calculator to check your answer.

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Example 3A: Solving ax2 + bx = c by Completing the Square


–3x2 + 12x – 15 = 0

Step 1

x2 – 4x + 5 = 0x2 – 4x = –5

x2 + (–4x) = –5

Step 3 x2 – 4x + 4 = –5 + 4

Divide by – 3 to make a = 1.



.Step 2

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Example 3A Continued


–3x2 + 12x – 15 = 0

Step 4 (x – 2)2 = –1

There is no real number whose square is negative, so there are no real solutions.


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Example 3B: Solving ax2 + bx = c by Completing the Square


5x2 + 19x = 4

Step 1 Divide by 5 to make a = 1.


Step 2 .

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Complete the square.Step 3



Factor and simplify.Step 4

Step 5Take the square root

of both sides.

Rewrite using like denominators.

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Step 6

The solutions are and –4.


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Check It Out! Example 3a


3x2 – 5x – 2 = 0



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Step 3

Step 4



Step 2 .

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Step 6

−


Step 5

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Check It Out! Example 3b


4t2 – 4t + 9 = 0



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Check It Out! Example 3b Continued


4t2 – 4t + 9 = 0

Step 2

Step 3



Step 4

There is no real number whose square is negative, so there are no real solutions.

.

Holt Algebra 1


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.

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2 Make a Plan

Example 4 Continued

Set the formula for the area of a rectangle equal to 195, the area of the room. Solve the equation.

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Example 4 Continued

Solve3

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

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Step 1 x2 + 2x = 195

Step 2

Step 3 x2 + 2x + 1 = 195 + 1

Step 4 (x + 1)2 = 196

Simplify.

Complete the square by adding 1 to both sides.

Factor the perfect-square trinomial.

Example 4 Continued


Step 5 x + 1 = ± 14

Step 6 x + 1 = 14 or x + 1 = –14 Write and solve two equations.x = 13 or x = –15

.

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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.

Example 4 Continued

Look Back4

The length of the room is 2 feet greater than the width. Also 13(15) = 195.

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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.

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2 Make a Plan

Set the formula for the area of a rectangle equal to 400, the area of the room. Solve the equation.

Check It Out! Example 4 Continued

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Solve3

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

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Step 1 x2 + 8x = 400

Step 3 x2 + 8x + 16 = 400 + 16

Step 4 (x + 4)2 = 416

Simplify.

Complete the square by adding 16 to both sides.

Factor the perfect-square trinomial.

Step 2



Step 5 x + 4 ± 20.4

Step 6 x + 4 20.4 or x + 4 –20.4Write and solve two

equations.x 16.4 or x –24.4

.

Holt Algebra 1


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.


Look Back4

The length of the room is 8 feet longer than the width. Also 16.4(24.4) = 400.16, which is approximately 400.

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1. x2 +11x +

2. x2 – 18x +


3. x2 – 2x – 1 = 0

4. 3x2 + 6x = 144

5. 4x2 + 44x = 23

Lesson Quiz: Part I

81

6, –8

Holt Algebra 1


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

holt algebra 1 9-8 completing the square warm up simplify. 19 1.2. 3.4

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