Review for Mastery
Name _______________________________________ Date __________________ Class__________________ LESSON
Review for Mastery
9-8
Completing the Square
You have already learned to solve quadratic equations by using square roots. This only works if the quadratic expression is a perfect square. Remember that perfect square trinomials can be written as perfect squares. x 2 + 8x + 16 = (x + 4)2
x 2 10x + 25 = (x 5)2 2
b If you have an equation of the form x + bx, you can add the term to make a perfect 2 square trinomial. This makes it possible to solve by using square roots. 2
Complete the square of x2 + 12x to form a perfect square trinomial. Then factor.
Complete the square of x2 + 7x to form a perfect square trinomial. Then factor.
x2 + 12x
x2 + 7x
Identify b.
2
2
Identify b.
2
12 2 2 = 6 = 36
b Find . 2
7 49 2 = 4
x2 + 12x + 36
b Add . 2
x2 + 7x +
Factor.
7 x + 2
2
2
(x + 6)
49 4
2
b Find . 2 2
b Add . 2
2
Factor.
Complete the square to form a perfect square trinomial by filling in the blanks. Then factor. 1. x2 14x
2. x2 + 20x
2
3. x2 + 6x
2
2
b 2 = _________________
b 2 = _________________
b 2 = _________________
x 2 14x +
x 2 + 20x +
x 2 + 6x +
(
)
2
(
)
2
(
)
2
Complete the square to form a perfect square trinomial. Then factor. 4. x2 + 18x
5. x2 16x
6. x2 + 5x
________________________
________________________
________________________
________________________
________________________
________________________
Original content Copyright © by Holt McDougal. Additions and changes to the original content are the responsibility of the instructor.
9-62
Holt McDougal Algebra 1
Name ________________________________________ Date __________________ Class__________________ LESSON
9-8
Review for Mastery Completing the Square continued
To solve a quadratic equation in the form x2 + bx = c, first complete the square of x2 + bx. Then solve using square roots. Solve x2 + 10x = −24 by completing the square. Step 1: Write equation in form x2 + bx = c. Identify b.
Step 4: Factor the perfect square trinomial on the left.
x2 + 10x = −24
x2 + 10x + 25 = 1 (x + 5)2 = 1 2
⎛ b⎞ Step 2: Find ⎜ ⎟ . ⎝ 2⎠
Step 5: Take the square root of both sides.
2
⎛ 10 ⎞ 2 ⎜⎝ 2 ⎟⎠ = 5 = 25
(x + 5)2 = ± 1 x + 5 = ±1 2
⎛ b⎞ Step 3: Add ⎜ ⎟ to both sides. ⎝ 2⎠
Step 6: Write and solve two equations.
x2 + 10x = −24 +25
x+5=1
+25
OR
−5 −5
2
x + 10x + 25 = 1
x = −4
x + 5 = −1 −5 −5
OR
x = −6
The solutions are −4 and −6. Solve by completing the square.
7. x2 − 6x = 7
8. x2 + 8x = −12
_________________________________________ 2
________________________________________
10. x2 + 4x − 32= 0
9. x − 2x − 63 = 0
_________________________________________
________________________________________
Original content Copyright © by Holt McDougal. Additions and changes to the original content are the responsibility of the instructor.
9-63
Holt McDougal Algebra 1
LESSON 9–9
Review for Mastery 1. 49; 49; x 7
2. 100; 100; x + 10
3. 9; 9; x + 3 5. x2 16x + 64; (x 8)2 6. x2 + 5x +
Practice A 1. 1; 6; 5; 6; 6; 1; 5; 1; 1, 5
2
4. x + 18x + 81; (x + 9)2
5 25 ; x + 2 4
7. 7; 1
2. 1; 9; 20; 9; 9; 1; 20; 1; 5, 4 1 3. 2; 9; 4; , 4 4. 1; 3; 18; 6, 3 2
2
5. 3; 1; 5; 11; no real solutions
8. 2, 6
9. 9; 7
10. 4; 8
Challenge
7. 64; 2
8. 8, 8
9. 6
10. 4, 8 2
5 1. y = x + 1 2 2 5 4 2
b 3. y = x + + c 2
6. 10; 1; 25; 0; 1
2
11 2. y = x + 6 2 2 11 4
1. 3, 4
b b2 4. ,c 4 2
3. 3,
5. a > 0
1 2
2. 5, 4.
3 4
1 , 5 3
5. no real solutions
6. 2
7. 1
8. 3, 5
9. 7, 7
6. If the parabola opens upward and the y-coordinate of the vertex is greater than 0, then the parabola is completely above the x-axis. If it is less than 0, then the parabola crosses the x-axis at two points. If it is equal to 0, then the parabola touches the x-axis at one point.
10.
1 1 , 3 2
11. 10, 2 12. No; the discriminant is negative so it will never reach the given height. Practice C
8. a. (1, 9) b. 2
9. a. (4.5, 2.25) b. 2 Problem Solving 1. 11 ft by 16 ft
1 , 5 2
Practice B
b2 4
7. a. 0 b. 1 c. 2
11.
1. 11, 4
2.
1 2 , 4 3
3. 15, 15
4.
1 5 , 2 4
5. 2
6. no real solutions
7. 1
8. no real solutions
9.
2. 2.5 ft
3. 38 in. by 28 in.; 1064 square in.
5 , 4 2
10. 6, 6
2 3
4. C
5. G
11.
6. B
7. J
12. For the 2 in., 3 in., 4 in., and 5 in. shells, the discriminant is negative. The discriminant for the 6 in. shell is positive, so that is the smallest shell that will reach over the monument.
Reading Strategies 1. x = 10 or x = 6
2. x = 9 or x = 19
3. x = 3 or x = 13 4. x = 2
1 1 or x = 3 2 2
Original content Copyright © by Holt McDougal. Additions and changes to the original content are the responsibility of the instructor.
A35
Holt McDougal Algebra 1
Review for Mastery
9-8
Completing the Square
You have already learned to solve quadratic equations by using square roots. This only works if the quadratic expression is a perfect square. Remember that perfect square trinomials can be written as perfect squares. x 2 + 8x + 16 = (x + 4)2
x 2 10x + 25 = (x 5)2 2
b If you have an equation of the form x + bx, you can add the term to make a perfect 2 square trinomial. This makes it possible to solve by using square roots. 2
Complete the square of x2 + 12x to form a perfect square trinomial. Then factor.
Complete the square of x2 + 7x to form a perfect square trinomial. Then factor.
x2 + 12x
x2 + 7x
Identify b.
2
2
Identify b.
2
12 2 2 = 6 = 36
b Find . 2
7 49 2 = 4
x2 + 12x + 36
b Add . 2
x2 + 7x +
Factor.
7 x + 2
2
2
(x + 6)
49 4
2
b Find . 2 2
b Add . 2
2
Factor.
Complete the square to form a perfect square trinomial by filling in the blanks. Then factor. 1. x2 14x
2. x2 + 20x
2
3. x2 + 6x
2
2
b 2 = _________________
b 2 = _________________
b 2 = _________________
x 2 14x +
x 2 + 20x +
x 2 + 6x +
(
)
2
(
)
2
(
)
2
Complete the square to form a perfect square trinomial. Then factor. 4. x2 + 18x
5. x2 16x
6. x2 + 5x
________________________
________________________
________________________
________________________
________________________
________________________
Original content Copyright © by Holt McDougal. Additions and changes to the original content are the responsibility of the instructor.
9-62
Holt McDougal Algebra 1
Name ________________________________________ Date __________________ Class__________________ LESSON
9-8
Review for Mastery Completing the Square continued
To solve a quadratic equation in the form x2 + bx = c, first complete the square of x2 + bx. Then solve using square roots. Solve x2 + 10x = −24 by completing the square. Step 1: Write equation in form x2 + bx = c. Identify b.
Step 4: Factor the perfect square trinomial on the left.
x2 + 10x = −24
x2 + 10x + 25 = 1 (x + 5)2 = 1 2
⎛ b⎞ Step 2: Find ⎜ ⎟ . ⎝ 2⎠
Step 5: Take the square root of both sides.
2
⎛ 10 ⎞ 2 ⎜⎝ 2 ⎟⎠ = 5 = 25
(x + 5)2 = ± 1 x + 5 = ±1 2
⎛ b⎞ Step 3: Add ⎜ ⎟ to both sides. ⎝ 2⎠
Step 6: Write and solve two equations.
x2 + 10x = −24 +25
x+5=1
+25
OR
−5 −5
2
x + 10x + 25 = 1
x = −4
x + 5 = −1 −5 −5
OR
x = −6
The solutions are −4 and −6. Solve by completing the square.
7. x2 − 6x = 7
8. x2 + 8x = −12
_________________________________________ 2
________________________________________
10. x2 + 4x − 32= 0
9. x − 2x − 63 = 0
_________________________________________
________________________________________
Original content Copyright © by Holt McDougal. Additions and changes to the original content are the responsibility of the instructor.
9-63
Holt McDougal Algebra 1
LESSON 9–9
Review for Mastery 1. 49; 49; x 7
2. 100; 100; x + 10
3. 9; 9; x + 3 5. x2 16x + 64; (x 8)2 6. x2 + 5x +
Practice A 1. 1; 6; 5; 6; 6; 1; 5; 1; 1, 5
2
4. x + 18x + 81; (x + 9)2
5 25 ; x + 2 4
7. 7; 1
2. 1; 9; 20; 9; 9; 1; 20; 1; 5, 4 1 3. 2; 9; 4; , 4 4. 1; 3; 18; 6, 3 2
2
5. 3; 1; 5; 11; no real solutions
8. 2, 6
9. 9; 7
10. 4; 8
Challenge
7. 64; 2
8. 8, 8
9. 6
10. 4, 8 2
5 1. y = x + 1 2 2 5 4 2
b 3. y = x + + c 2
6. 10; 1; 25; 0; 1
2
11 2. y = x + 6 2 2 11 4
1. 3, 4
b b2 4. ,c 4 2
3. 3,
5. a > 0
1 2
2. 5, 4.
3 4
1 , 5 3
5. no real solutions
6. 2
7. 1
8. 3, 5
9. 7, 7
6. If the parabola opens upward and the y-coordinate of the vertex is greater than 0, then the parabola is completely above the x-axis. If it is less than 0, then the parabola crosses the x-axis at two points. If it is equal to 0, then the parabola touches the x-axis at one point.
10.
1 1 , 3 2
11. 10, 2 12. No; the discriminant is negative so it will never reach the given height. Practice C
8. a. (1, 9) b. 2
9. a. (4.5, 2.25) b. 2 Problem Solving 1. 11 ft by 16 ft
1 , 5 2
Practice B
b2 4
7. a. 0 b. 1 c. 2
11.
1. 11, 4
2.
1 2 , 4 3
3. 15, 15
4.
1 5 , 2 4
5. 2
6. no real solutions
7. 1
8. no real solutions
9.
2. 2.5 ft
3. 38 in. by 28 in.; 1064 square in.
5 , 4 2
10. 6, 6
2 3
4. C
5. G
11.
6. B
7. J
12. For the 2 in., 3 in., 4 in., and 5 in. shells, the discriminant is negative. The discriminant for the 6 in. shell is positive, so that is the smallest shell that will reach over the monument.
Reading Strategies 1. x = 10 or x = 6
2. x = 9 or x = 19
3. x = 3 or x = 13 4. x = 2
1 1 or x = 3 2 2
Original content Copyright © by Holt McDougal. Additions and changes to the original content are the responsibility of the instructor.
A35
Holt McDougal Algebra 1
