Algebra unit 10.5

UNIT 10.5 SOLVING QUADRATICS UNIT 10.5 SOLVING QUADRATICS BY COMPLETING THE SQUAREBY COMPLETING THE SQUARE

Warm Up

Simplify.

19 1. 2.

3. 4.

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

Solve quadratic equations by completing the square.

Objective

completing the square

Vocabulary

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.

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 + B. x2 – 6x +

x2 + 2x

x2 + 2x + 1

x2 + –6x

x2 – 6x + 9

Identify b.

.

Check It Out! Example 1


a. x2 + 12x + b. x2 – 5x +

x2 + 12x

x2 + 12x + 36

x2 + –5xIdentify b.

x2 – 6x +

.

Check It Out! Example 1


c. 8x + x2 +

x2 + 8x

x2 + 12x + 16

Identify b.

.

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

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.

.

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 – 16 –15–15 –15

x2 + 16x = –15

(–15)2 + 16(–15) –15

225 – 240 –15–15 –15

Example 2B: Solving x2 +bx = c

Solve by completing the square.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

Example 2B Continued


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

Check Use a graphing calculator to check your answer.

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




.

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

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

.

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.

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

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.


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.


Step 2 .

Complete the square.Step 3



Factor and simplify.Step 4

Step 5Take the square root

of both sides.

Rewrite using like denominators.



Step 6

The solutions are and –4.


Check It Out! Example 3a

Solve by completing the square.3x2 – 5x – 2 = 0





Step 3

Step 4



Step 2 .




Step 6

−


Step 5

Check It Out! Example 3b

Solve by completing the square.4t2 – 4t + 9 = 0



Check It Out! Example 3b ContinuedSolve by completing the square.

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.

.

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.

11 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.

22 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.

Example 4 Continued

Solve33

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

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

.

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 Back44

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

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.

11 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.

22 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

Solve33

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

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

.

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 Back44

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


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

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

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Algebra unit 10.5

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Transcript of Algebra unit 10.5