Mscc9 wsk 0900 a - Schoolwires

Chapter 9

Chapter 9 Opener

Try It Yourself (p. 453) 1. 81 = 92 = 9

2. - 169 = - 132 = -13

3. ?

9 =? 25

32 = ? 3

52

5

4. - 6.25 = - 2.52 = - 2.5

5. 54 = 9 ? 6 = 9 ? 6 = 3 6 6. 80 = 16 ? 5 = 16 ? 5 = 4 5 7. 200 = 100 ? 2 = 100 ? 2 = 10 2

8. x2 + 10x + 25 = x2 + 2(x)(5) + 52 = (x + 5)2

9. m2 - 20m + 100 = m2 + 2(m)(-10) + (-10)2 = (m - 10)2

10. p2 + 12 p + 36 = p2 + 2( p)(6) + 62 = ( p + 6)2

Section 9.1

9.1 Activity (pp. 454 ? 455)

1. a.

y 10

9 8 7 6 5 4 3

-7 -6 -5 -4 -3 -2 O 1 3 4 5 6 7 x

-2 y = x2 - 2x -3 -4

b. The x-intercepts are points where the graph crosses the x-axis. In the graph of y = x2 - 2x, there are

2 x-intercepts. They are (0, 0) and (2, 0).

c. A solution of an equation in x is an x-value that makes the equation true. The equation x2 - 2x = 0 has two solutions. They are x = 0 and x = 2.

d. You can verify that the x-values found in part (c) are solutions of x2 - 2x = 0 by substituting the x-values into the left side of the equation and making sure it equals zero.

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2. a. x2 - 4 = 0

(-2, 0) -6 -5 -4 -3

y 6 5 4 3 2 1

O1

-2

(2, 0) 3 4 5 6x

-5 y = x2 - 4 -6

The solutions are x = - 2 and x = 2.

b. x2 + 3x = 0

(-3, 0) -6 -5 -4 -2

y 6 5 4 3 2 1 (0, 0)

1 2 3 4 5 6x

y = x2 + 3x -3 -4 -5 -6

The solutions are x = -3 and x = 0.

c. -x2 + 2x = 0

y 4

3 2 y = -x2 + 2x

(0, 0)

(2, 0)

-6 -5 -4 -3 -2

1 3 4 5 6x

-4 -5 -6 -7 -8

The solutions are x = 0 and x = 2.

Big Ideas Math Algebra 1 397 Worked-Out Solutions

Chapter 9

d. x2 - 2x + 1 = 0

y 9 8 7 6 5 4

y = x2 - 2x + 1 -6 -5 -4 -3 -2 O

(1, 0) 1 2 3 4 5 6x

-2 -3

The solution is x = 1.

3. You can get the equation equal to 0, set it equal to y, and then graph the resulting equation. The solutions of a quadratic equation in one variable are the x-intercepts of the graph.

4. You can substitute the solutions for the variable and make sure the equation is true.

Activity 2a. x = - 2 and x = 2

x = -2:

x = 2:

x2 - 4 = 0

(- 2)2

-

4

?

=

0

?

4-4 = 0

0=09

x2 - 4 = 0

?

22 - 4 = 0

?

4-4 = 0

0 = 09

Activity 2b. x = -3 and x = 0

x = -3:

x = 0:

x2 + 3x = 0

(- 3)2

+

3(- 3)

?

=

0

?

9-9 = 0

0 = 09

x2 + 3x = 0

?

02 + 3(0) = 0

?

0+0 = 0

0 = 09

Activity 2c. x = 0 and x = 2 x = 0:

x = 2:

- x2 + 2x = 0

?

- 02 + 2(0) = 0

?

-0 + 0 = 0

0 = 09

- x2 + 2x = 0

?

- 22 + 2(2) = 0

?

-4 + 4 = 0

0 = 09

398 Big Ideas Math Algebra 1 Worked-Out Solutions

Activity 2d. x = 1

x = 1:

x2 - 2x + 1 = 0

?

12 - 2(1) + 1 = 0

?

1-2+1 = 0

0 = 09

9.1 On Your Own (pp. 456 ? 458)

1. x2 - x - 2 = 0

(-1, 0) -4 -3 -2

y 6 5 4 3 2 1

1

(2, 0) 3 4 5 6x

-3 y = x2 - x - 2 -4

So, the solutions are x = -1 and x = 2.

Check: x = -1:

x = 2:

x2 - x - 2 = 0

(-1)2

-

(-1)

-

2

?

=

0

?

1+1- 2 = 0

0 = 09

x2 - x - 2 = 0

?

22 - 2 - 2 = 0

?

4-2-2 = 0

0 = 09

2. x2 + 7x + 10 = 0

y

(-2, 0) (-5, 0) -8 -7 -6 -4 -3

5 4 3 2 1

O 1 2x

-2 y = x2 + 7x + 10 -3

So, the solutions are x = -5 and x = - 2.

Check:

x = -5:

x2 + 7x + 10 = 0

(-5)2

+

7(-5)

+ 10

?

=

0

?

25 + 7(-5) + 10 = 0

?

25 - 35 + 10 = 0

0 = 09

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Chapter 9

x = -2:

x2 + 7x + 10 = 0

(- 2)2

+

7(- 2)

+ 10

?

=

0

?

4 + 7(-2) + 10 = 0

?

4 - 14 + 10 = 0

0 = 09

3. x2 + x = 12

(-4, 0) -8 -6

y 6

4

2

(3, 0)

O 2 6 8 10 12 x

-4 -6 -8

-14 y = x2 + x - 12

So, the solutions are x = - 4 and x = 3.

Check: x = -4:

x = 3:

x2 + x = 12

(- 4)2

+

(- 4)

?

=

12

?

16 - 4 = 12

12 = 12 9

x2 + x = 12

?

32 + 3 = 12

?

9 + 3 = 12

12 = 12 9

4. x2 + 1 = 2x

y 9 8 7 6 5 4

y = x2 - 2x + 1 -6 -5 -4 -3 -2 O

(1, 0) 4 x

So, the solution is x = 1.

Check:

x = 1: x2 + 1 = 2x

?

12 + 1 = 2(1)

?

1+1 = 2

2 = 29

5. x2 + 4x = 0

(-4, 0) -7 -6 -5 -3 -2

y 5 4 3 2 1 (0, 0)

1 2 3x

-4 y = x2 + 4x -5

So, the solutions are x = -4 and x = 0.

Check: x = -4:

x = 0:

x2 + 4x = 0

(-4)2

+

4(- 4)

?

=

0

?

16 + 4(- 4) = 0

?

16 - 16 = 0

0 = 09

x2 + 4x = 0

?

02 + 4(0) = 0

?

0 + 4(0) = 0

?

0+0 = 0

0 = 09

6. x2 + 10x = - 25

-9 -8 -7

y 8 7 6 5 4 3 2 (-5, 0) 1 -5 -4 -3 -2 O 1 x

y = x2 + 10x + 25 -2

So, the solution is x = -5.

Check:

x = -5:

x2 + 10x = - 25

(-5)2

+ 10(-5)

?

=

- 25

?

25 + 10(-5) = -25

?

25 - 50 = -25

-25 = -25 9

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Big Ideas Math Algebra 1 399 Worked-Out Solutions

Chapter 9

7. x2 = 3x - 3 y = x2 y = 3x - 3

y 5

4

3

2 y = x2 1

-5 -4 -3 -2 O

2 3 4 5x

-2 -3 y = 3x - 3

The graphs do not intersect. So, x2 = 3x - 3 has no real solutions.

8. x2 + 7x = - 6

y

(-6, 0)

(-1, 0) 1

-9 -8 -7 -5 -4 -3 -2 O 1 x

-2

-3

-4

-5

-6

y = x2 + 7x + 6

-7

So, the solutions are x = - 6 and x = -1.

Check: x = -6:

x = -1:

x2 + 7x = -6

(- 6)2

+

7(-6)

?

=

-6

?

36 + 7(-6) = -6

?

36 - 42 = -6

-6 = -6 9

x2 + 7x = -6

(-1)2

+

7(-1)

?

=

-6

?

1 + 7(-1) = -6

?

1 - 7 = -6

-6 = -6 9

9. 2x + 5 = - x2 y = 2x + 5

y = - x2

y 5 y = 2x + 5 3 2 1

-5 -4

-2 -3 -4 -5

2 3 4 5x y = -x2

The graphs do not intersect. So, 2x + 5 = - x2 has no real solutions.

400 Big Ideas Math Algebra 1 Worked-Out Solutions

10. -16t2 + 75t + 2 = 65 - 65 - 65

-16t2 + 75t - 63 = 0

80 y = -16t 2 + 75t - 63

-1

5

-10

Use the zero feature to find that the football is 65 feet above the ground after about 1.1 seconds and about 3.6 seconds.

9.1 Exercises (pp. 459 ? 461)

Vocabulary and Concept Check 1. A quadratic equation is a nonlinear equation that can be written in the standard form ax2 + bx + c = 0, where a 0.

2. The equation x2 + x - 4 = 0 does not belong. It is the only equation in standard form.

3. You can use a graph to find the number of solutions of a quadratic equation by finding the x-intercepts.

4. The roots, or solutions, of an equation are the same as the zeros of the related function or the x-intercepts of its graph.

Practice and Problem Solving

5. x2 - 10x + 24 = 0

The solutions are x = 4 and x = 6.

Check:

x = 4:

x = 6:

x2 - 10x + 24 = 0

?

42 - 10(4) + 24 = 0

?

16 - 10(4) + 24 = 0

?

16 - 40 + 24 = 0

0=09

x2 - 10x + 24 = 0

?

62 - 10(6) + 24 = 0

?

36 - 10(6) + 24 = 0

?

36 - 60 + 24 = 0

0 = 09

6. - x2 - 4x - 6 = 0

The graph does not intersect the x-axis. So, - x2 - 4x - 6 = 0 has no real solutions.

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Chapter 9

7. x2 + 12x + 36 = 0

The solution is x = -6.

Check:

x = -6:

x2 + 12x + 36 = 0

(- 6)2

+ 12(-6)

+

36

?

=

0

?

36 + 12(-6) + 36 = 0

?

36 - 72 + 36 = 0

0 = 09

8. x2 - 4x = 0

y 5 4

(0, 0)

(4, 0)

-3 -2 O 1 2 3 5 6 7 x

-2

-3

-4

-5

y = x2 - 4x

So, the solutions are x = 0 and x = 4.

Check: x = 0:

x = 4:

x2 - 4x = 0

?

02 - 4(0) = 0

?

0 - 4(0) = 0

?

0-0 = 0

0 = 09

x2 - 4x = 0

?

42 - 4(4) = 0

?

16 - 4(4) = 0

?

16 - 16 = 0

0 = 09

9. x2 - 6x + 9 = 0

y 8

7

6

5

4

3

(3, 0)

2

1

-2 O 1 2 3 5 6 7 8 x

-2

y = x2 - 6x + 9

So, the solution is x = 3.

Check:

x = 3: x2 - 6x + 9 = 0

?

32 - 6(3) + 9 = 0

?

9 - 6(3) + 9 = 0

?

9 - 18 + 9 = 0

0=09

10. x2 - 6x - 7 = 0

(-1, 0) -6 -4

y (7, 0)

2 4 6 8 10 12 14 x

-6

-8

-10

-12

-14

-16

-18

y = x2 - 6x - 7

So, the solutions are x = -1 and x = 7.

Check: x = -1:

x = 7:

x2 - 6x - 7 = 0

(-1)2

-

6(-1)

-

7

?

=

0

?

1 - 6(-1) - 7 = 0

?

1+6-7 = 0

0=09

x2 - 6x - 7 = 0

?

72 - 6(7) - 7 = 0

?

49 - 6(7) - 7 = 0

?

49 - 42 - 7 = 0

0 = 09

11. x2 - 2x + 5 = 0

y 9 8

-4 -3 -2

5 4 3 y = x2 - 2x + 5 2 1

O 1 2 3 4 5 6x

The graph does not intersect the x-axis. So, x2 - 2x + 5 = 0 has no real solutions.

Copyright ? Big Ideas Learning, LLC All rights reserved.

Big Ideas Math Algebra 1 401 Worked-Out Solutions

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