b. Make a table of values that includes the
vertex.
4-1 Graphing Quadratic Functions
c. Use this information to graph the function.
Complete parts a–c for each quadratic function.
13. a. Find the y-intercept, the equation of the axis
of symmetry, and the x-coordinate of the
vertex.
SOLUTION: a. Compare the function
with the standard form of a quadratic function.
b. Make a table of values that includes the
vertex.
Here, a = –2, b = 0 and c = 0.
c. Use this information to graph the function.
The y-intercept is 0.
13. The equation of the axis of symmetry is
SOLUTION: a. Compare the function
with the standard form of a quadratic function.
.
Therefore, x = 0 is the axis of symmetry.
The x-coordinate of the vertex is
Here, a = –2, b = 0 and c = 0.
.
b. Substitute –2, –1, 0, 1 and 2 for x and make the
table.
The y-intercept is 0.
The equation of the axis of symmetry is
.
Therefore, x = 0 is the axis of symmetry.
The x-coordinate of the vertex is
.
c. Graph the function.
b. Substitute –2, –1, 0, 1 and 2 for x and make the
table.
c. Graph the function.
15. eSolutions Manual - Powered by Cognero
SOLUTION: Page 1
a. Compare the function
with the standard form of a quadratic function.
4-1 Graphing
Quadratic Functions
15. 17. SOLUTION: SOLUTION: a. Compare the function
with the standard form of a quadratic function.
Here, a = 4, b = 0 and c = –3.
a. Compare the function
standard form of a quadratic function.
Here, a = –3, b = 0 and c = 5.
The y-intercept is –3.
The y-intercept is 5.
The equation of the axis of symmetry is
with the .
The equation of the axis of symmetry is
.
Therefore, x = 0 is the axis of symmetry.
Therefore, x = 0 is the equation of axis of symmetry.
The x-coordinate of the vertex is
.
The x-coordinate of the vertex is
.
b. Substitute –2, –1, 0, 1 and 2 for x and make the
table.
b. Substitute –2, –1, 0, 1 and 2 for x and make the
table.
c. Graph the function.
c. Graph the function
.
17. 19. SOLUTION: a. Compare the function
standard form of a quadratic function.
SOLUTION: with the a. Compare the function
standard form of a quadratic function.
Here, a = 1, b = –3 and c = –10.
Here, a = –3, b = 0 and c = 5.
eSolutions Manual - Powered by Cognero
with the
Page 2
4-1 Graphing Quadratic Functions
21. 19. SOLUTION: SOLUTION: a. Compare the function
standard form of a quadratic function.
Here, a = 1, b = –3 and c = –10.
a. Compare the function
the standard form of a quadratic function.
Here, a = –2, b = 3 and c = 9.
The y-intercept is 9.
The equation of the axis of symmetry is
with the
The y-intercept is –10.
.
The equation of the axis of symmetry is
The equation of the axis of symmetry is x = 0.75.
.
The x-coordinate of the vertex is
Therefore, x = 1.5 is the equation of the axis of
symmetry.
The x-coordinate of the vertex is
with .
b. Substitute –1, 0, 0.75, 1.5 and 2.5 for x and make
the table.
.
b. Substitute 0, 1, 1.5, 2 and 3 for x and make the
table.
c. Graph the function.
c. Graph the function.
Determine whether each function has a
maximum or minimum value, and find that
value. Then state the domain and range of the
function.
23. SOLUTION: 21. Compare the function
with the standard form of a quadratic function.
SOLUTION: a. Compare
the function
eSolutions
Manual - Powered
by Cognero
the standard form of a quadratic function.
Here, a = –2, b = 3 and c = 9.
with Here, a = –1, b = 0 and c = –12.
Page 3
For this function, a = –1, so the graph opens down
The range is all real numbers less than or equal to
the maximum value.
4-1 Graphing Quadratic Functions
Determine whether each function has a
maximum or minimum value, and find that
value. Then state the domain and range of the
function.
25. SOLUTION: Compare the function
standard form of a quadratic function.
23. with the Here, a = –1, b = –7 and c = 1.
SOLUTION: Compare the function
with the standard form of a quadratic function.
For this function, a = –1, so the graph opens down
and the function has a maximum value.
Here, a = –1, b = 0 and c = –12.
The x-coordinate of the vertex is
.
For this function, a = –1, so the graph opens down
and the function has a maximum value.
Substitute –3.5 for x in the function to find the ycoordinate of the vertex.
The x-coordinate of the vertex is
.
Substitute 0 for x in the function to find the ycoordinate of the vertex.
Therefore, the maximum value of the function is
13.25.
Therefore, the maximum value of the function is –
12.
The domain is all real numbers.
D = {all real numbers}.
The range is all real numbers less than or equal to
the maximum value.
The domain is all real numbers.
D = {all real numbers}.
The range is all real numbers less than or equal to the
maximum value.
27. SOLUTION: Compare the function
standard form of a quadratic function.
25. SOLUTION: Here, a = –2, b = –4 and c = 5.
Compare the function
standard form of a quadratic function.
with the For this function, a = –2, so the graph opens down
and the function has a maximum value.
Here, a = –1, b = –7 and c = 1.
The x-coordinate of the vertex is
For this function, a = –1, so the graph opens down
and the
function
hasbya Cognero
maximum value.
eSolutions
Manual
- Powered
The x-coordinate of the vertex is
with the .
Substitute –1 for x in the function to find the ycoordinate of the vertex.
Page 4
The range is all real numbers less than or equal to
the maximum value.
The range is all real numbers less than or equal to the
maximum value.
4-1 Graphing Quadratic Functions
27. 29. SOLUTION: SOLUTION: Compare the function
standard form of a quadratic function.
with the Compare the function
standard form of a quadratic function.
with the Here, a = –2, b = –4 and c = 5.
Here, a = 1, b = 12 and c = 27.
For this function, a = –2, so the graph opens down
and the function has a maximum value.
For this function, a = 1, so the graph opens up and
the function has a minimum value.
The x-coordinate of the vertex is
The x-coordinate of the vertex is
.
.
Substitute –6 for x in the function to find the ycoordinate of the vertex.
Substitute –1 for x in the function to find the ycoordinate of the vertex.
Therefore, the minimum value of the function is –9.
Therefore, the maximum value of the function is 7.
The domain is all real numbers.
D = {all real numbers}.
The range is all real numbers less than or equal to
the maximum value.
The domain is all real numbers.
D = {all real numbers}.
The range is all real numbers greater than or equal to
the minimum value.
31. 29. SOLUTION: Compare the function
the standard form of a quadratic function.
SOLUTION: Compare the function
standard form of a quadratic function.
with the For this function, a = 1, so the graph opens up and
the function has a minimum value.
The x-coordinate of the vertex is
The x-coordinate of the vertex is
.
Here, a = 2, b = –16 and c = –42.
For this function, a = 2, so the graph opens up and
the function has a minimum value.
Here, a = 1, b = 12 and c = 27.
coordinate of the vertex.
eSolutions
Manual –6
- Powered
by the
Cognero
Substitute
for x in
function
with to find the y-
.
Substitute 4 for x in the function to find the ycoordinate of the vertex.
Page 5
The range is all real numbers greater than or equal to
the minimum value.
4-1 Graphing Quadratic Functions
31. SOLUTION: Compare the function
the standard form of a quadratic function.
with Here, a = 2, b = –16 and c = –42.
For this function, a = 2, so the graph opens up and
the function has a minimum value.
The x-coordinate of the vertex is
.
Substitute 4 for x in the function to find the ycoordinate of the vertex.
Therefore, the minimum value of the function is –74.
The domain is all real numbers.
D = {all real numbers}.
The range is all real numbers greater than or equal to
the minimum value.
.
eSolutions Manual - Powered by Cognero
Page 6
b. Make a table of values that includes the
vertex.
4-1 Graphing Quadratic Functions
c. Use this information to graph the function.
Complete parts a–c for each quadratic function.
12. a. Find the y-intercept, the equation of the axis
of symmetry, and the x-coordinate of the
vertex.
SOLUTION: Here, a = 4, b = 0 and c = 0.
b. Make a table of values that includes the
vertex.
The y-intercept is 0.
c. Use this information to graph the function.
The equation of the axis of symmetry is
12. .
Therefore, x = 0 is the axis of symmetry.
SOLUTION: The x-coordinate of the vertex is
Here, a = 4, b = 0 and c = 0.
.
b. Substitute –2, –1, 0, 1 and 2 for x and make the
table.
The y-intercept is 0.
The equation of the axis of symmetry is
.
Therefore, x = 0 is the axis of symmetry.
The x-coordinate of the vertex is
.
c. Graph the function.
b. Substitute –2, –1, 0, 1 and 2 for x and make the
table.
c. Graph the function.
16. SOLUTION: eSolutions Manual - Powered by Cognero
a. Compare the function
with the Page 1
standard form of a quadratic function.
Here, a = 1, b = 0 and c = 3.
4-1 Graphing
Quadratic Functions
16. 20. SOLUTION: SOLUTION: a. Compare the function
the standard form of a quadratic function.
Here, a = –1, b = 4 and c = –6.
The y-intercept is –6.
The equation of the axis of symmetry is
a. Compare the function
with the standard form of a quadratic function.
Here, a = 1, b = 0 and c = 3.
The y-intercept is 3.
.
The equation of the axis of symmetry is
Therefore, x = 2 is the equation of the axis of
symmetry.
.
The x-coordinate of the vertex is
Therefore, x = 0 is the axis of symmetry.
The x-coordinate of the vertex is
with .
b. Substitute 0, 1, 2, 3 and 4 for x and make the table.
.
b. Substitute –2, –1, 0, 1 and 2 for x and make the
table.
c. Graph the function.
c. Graph the function.
Determine whether each function has a
maximum or minimum value, and find that
value. Then state the domain and range of the
function.
24. SOLUTION: 20. Compare the function
standard form of a quadratic function.
SOLUTION: eSolutions Manual - Powered by Cognero
a. Compare the function
the standard form of a quadratic function.
Here, a = –1, b = 4 and c = –6.
with Here, a = 1, b = –6 and c = 9.
with the Page 2
For this function, a = 1, so the graph opens up and
The range is all real numbers greater than or equal
to the minimum value.
4-1 Graphing Quadratic Functions
Determine whether each function has a
maximum or minimum value, and find that
value. Then state the domain and range of the
function.
28. SOLUTION: Compare the function
with the standard form of a quadratic function.
24. Here, a = –5, b = 0 and c = 15.
SOLUTION: Compare the function
standard form of a quadratic function.
with the For this function, a = –5, so the graph opens down
and the function has a maximum value.
Here, a = 1, b = –6 and c = 9.
The x-coordinate of the vertex is
.
For this function, a = 1, so the graph opens up and
the function has a minimum value.
The x-coordinate of the vertex is
Substitute 3 for x in the function to find the ycoordinate of the vertex.
Therefore, the minimum value of the function is 0.
The domain is all real numbers.
D = {all real numbers}.
The range is all real numbers greater than or equal
to the minimum value.
Substitute 0 for x in the function to find the ycoordinate of the vertex.
.
Therefore, the maximum value of the function is 15.
The domain is all real numbers.
D = {all real numbers}.
The range is all real numbers less than or equal to
the maximum value.
32. CCSS MODELING A financial analyst determined
that the cost, in thousands of dollars, of producing
2
bicycle frames is C = 0.000025f – 0.04f + 40, where
f is the number of frames produced.
a. Find the number of frames that minimizes cost.
28. SOLUTION: Compare the function
with the standard form of a quadratic function.
b. What is the total cost for that number of frames?
SOLUTION: a. The x-coordinate of the vertex is:
Here, a = –5, b = 0 and c = 15.
For this function, a = –5, so the graph opens down
value.
eSolutions
Manual
- Powered
and the
function
hasbya Cognero
maximum
The x-coordinate of the vertex is
The number of frames that minimize the cost is 800.
b. Substitute 800 for f in the function and simplify.
Page 3
The range is all real numbers less than or equal to
the maximum value.
4-1 Graphing Quadratic Functions
32. CCSS MODELING A financial analyst determined
that the cost, in thousands of dollars, of producing
2
bicycle frames is C = 0.000025f – 0.04f + 40, where
f is the number of frames produced.
a. Find the number of frames that minimizes cost.
b. What is the total cost for that number of frames?
SOLUTION: a. The x-coordinate of the vertex is:
The number of frames that minimize the cost is 800.
b. Substitute 800 for f in the function and simplify.
Therefore, the total cost is $24, 000.
eSolutions Manual - Powered by Cognero
Page 4
SOLUTION: The graph has no x-intercepts. Thus, the equation
has no real solution.
4-2 Solving Quadratic Equations by Graphing
Use the related graph of each equation to
determine its solutions.
16. 14. SOLUTION: The x-intercept of the graph is 2. Thus, the solution
of the equation is 2.
SOLUTION: The x-intercepts of the graph are –4 and 0. Thus, the
solutions of the equation are –4 and 0.
17. 15. SOLUTION: The x-intercept of the graph is –2. Thus, the solution
of the equation is –2.
SOLUTION: The graph has no x-intercepts. Thus, the equation
has no real solution.
18. 16. SOLUTION: The graph has no x-intercepts. Thus, the equation
has no real solution.
SOLUTION: The x-intercept of the graph is 2. Thus, the solution
of the equation is 2.
eSolutions Manual - Powered by Cognero
17. 19. Page 1
SOLUTION: The graph has no x-intercepts. Thus, the equation
has no real
solution.Equations by Graphing
4-2 Solving
Quadratic
The x-intercepts of the graph indicate that the
solutions are 0 and 5.
21. 19. SOLUTION: Graph the related function
.
SOLUTION: The x-intercepts of the graph are –3 and 4. Thus, the
solutions of the equation are –3 and 4.
The x-intercepts of the graph indicate that the
solutions are –2 and 0.
Solve each equation. If exact roots cannot be
found, state the consecutive integers between
which the roots are located.
22. 20. SOLUTION: SOLUTION: Graph the related function
Graph the related function
.
The x-intercepts of the graph indicate that the
solutions are –2 and 7.
The x-intercepts of the graph indicate that the
solutions are 0 and 5.
23. SOLUTION: Graph the related function
.
21. SOLUTION: Graph the related function
.
eSolutions Manual - Powered by Cognero
Page 2
The x-intercepts of the graph indicate that the
solutionsQuadratic
are –2 andEquations
7.
4-2 Solving
by Graphing
The x-intercept of the graph indicates that the
solution is 9.
25. 23. SOLUTION: SOLUTION: Graph the related function
.
Graph the related function
.
The x-intercepts of the graph indicate that the
solutions are –4 and 6.
24. SOLUTION: The graph has no x-intercepts. Thus, the equation
has no real solution.
26. Graph the related function
.
SOLUTION: Graph the related function
.
The x-intercept of the graph indicates that the
solution is 9.
25. SOLUTION: The x-intercepts of the graph indicate that one
solution is between –3 and –2, and the other solution
is between 3 and 4.
eSolutions Manual - Powered by Cognero
Graph the related function
Page 3
27. .
The x-intercepts of the graph indicate that one
solution is between 1 and 2, and the other solution is
between –1 and 0.
The x-intercepts of the graph indicate that one
solution is between –3 and –2, and the other solution
is between
3 and 4. Equations by Graphing
4-2 Solving
Quadratic
28. 27. SOLUTION: SOLUTION: Graph the equation
Graph the related function
.
.
The x-intercepts of the graph indicate that one
solution is between 10 and 11, and the other solution
is between –1 and 0.
The x-intercepts of the graph indicate that one
solution is between 1 and 2, and the other solution is
between –1 and 0.
29. 28. SOLUTION: SOLUTION: Graph the related function
Graph the related function
.
.
The x-intercepts of the graph indicate that one
solution is between 10 and 11, and the other solution
is between –1 and 0.
The graph has no x-intercepts. Thus, the equation
has no real solution.
eSolutions Manual - Powered by Cognero
29. Page 4
Therefore,
4-3 Solving Quadratic Equations by Factoring
Factor each polynomial.
23. 20. SOLUTION: Factor out 3.
SOLUTION: The GCF of the two terms is 8a. Factor the GCF.
2
2
2
2
2
2
Use the identity a – b = (a + b)(a – b) to factor x
– 4.
21. Therefore,
SOLUTION: The GCF of the two terms is 17c. Factor the GCF.
24. SOLUTION: Factor out 15.
22. SOLUTION: Factor 8x from the first two terms and –3a from the
last two terms.
Use the identity a – b = (a + b)(a – b) to factor y
– 16.
Therefore,
Factor 4y + 5b from the two terms.
25. Therefore,
SOLUTION: Factor 12c from the first two terms and –d from the
last two terms.
23. Factor 4g + 3f from the two terms.
SOLUTION: Factor out 3.
eSolutions Manual - Powered by Cognero
Page 1
Therefore,
Therefore,
Therefore,
4-3 Solving Quadratic Equations by Factoring
25. 27. SOLUTION: Factor 12c from the first two terms and –d from the
last two terms.
SOLUTION: Find the factors of –22 whose sum is –9.
2(–11) = –22 and 2 + (–11) = –9
Write –9x as 2x –11x.
Factor 4g + 3f from the two terms.
Factor x from the first two terms and –11 from the
last two terms.
Therefore,
Factor x + 2 from the two terms.
.
Therefore,
26. SOLUTION: Find the factors of 40 whose sum is 13.
28. SOLUTION: Here, a = 3, b = 12 and c = –36.
Write 13x as 5x + 8x.
ac = 3(–36) = –108
5(8) = 40 and 5 + 8 = 13
Find two factors of –108 whose sum is 12.
Factor x from the first two terms and 8 from the last
two terms.
–6(18) = –108 and –6 + 18 = 12
Write 12x as –6x + 18x.
Factor x + 5 from the two terms.
Factor 3x from the first two terms and 18 from the
last two terms.
Therefore,
Factor x – 2 from the two terms.
27. SOLUTION: eSolutions
- Powered
by Cognero
FindManual
the factors
of –22
whose
sum is –9.
2(–11) = –22 and 2 + (–11) = –9
Page 2
Therefore,
Therefore,
4-3 Solving Quadratic Equations by Factoring
28. 29. SOLUTION: Here, a = 3, b = 12 and c = –36.
SOLUTION: Here, a = 15, b = 7 and c = –2.
ac = 3(–36) = –108
ac = 15(–2) = –30
Find two factors of –108 whose sum is 12.
–6(18) = –108 and –6 + 18 = 12
Find two factors of –30 whose sum is 7.
10(–3) = –30 and 10 + (–3) = 7
Write 12x as –6x + 18x.
Write 7x as 10x – 3x.
Factor 3x from the first two terms and 18 from the
last two terms.
Factor 5x from the first two terms and –1 from the
last two terms.
Factor 3x + 2 from the two terms.
Factor x – 2 from the two terms.
Therefore,
Therefore,
30. SOLUTION: Here, a = 4, b = 29 and c = 30.
29. ac = 4(30) = 120
SOLUTION: Here, a = 15, b = 7 and c = –2.
Find two factors of 120 whose sum is 29.
5(24) = 120 and 5 + 24 = 29
ac = 15(–2) = –30
Write 29x as 5x + 24x.
Find two factors of –30 whose sum is 7.
10(–3) = –30 and 10 + (–3) = 7
Factor x from the first two terms and 6 from the last
two terms.
Write 7x as 10x – 3x.
eSolutions Manual - Powered by Cognero
Factor 5x from the first two terms and –1 from the
last two terms.
Factor 4x + 5 from the two terms.
Page 3
Therefore,
Therefore,
4-3 Solving Quadratic Equations by Factoring
30. 31. SOLUTION: Here, a = 4, b = 29 and c = 30.
SOLUTION: Here, a = 18, b = 15 and c = –12.
ac = 4(30) = 120
ac = 18(–12) = –216
Find two factors of 120 whose sum is 29.
Find two factors of –216 whose sum is 15.
5(24) = 120 and 5 + 24 = 29
24(–9) = –216 and 24 + (–9) = 15
Write 29x as 5x + 24x.
Write 15x as 24x + (–9)x.
Factor x from the first two terms and 6 from the last
two terms.
Factor 6x from the first two terms and –3 from the
last two terms.
Factor 4x + 5 from the two terms.
Factor 3x + 4 from the two terms
.
Therefore,
Therefore,
31. SOLUTION: Here, a = 18, b = 15 and c = –12.
ac = 18(–12) = –216
32. SOLUTION: 2
Factor z from all the three terms.
Find two factors of –216 whose sum is 15.
24(–9) = –216 and 24 + (–9) = 15
2
Factor 8x – 4x – 12.
Write 15x as 24x + (–9)x.
Here, a = 8, b = –4 and c = –12.
ac = 8(–12) = –96
Factor 6x from the first two terms and –3 from the
last two terms.
eSolutions Manual - Powered by Cognero
Find two factors of –96 whose sum is –4.
–12(8) = –96 and –12 + 8 = –4
Write –4x as –12x + 8x.
Page 4
Therefore,
4-3 Solving Quadratic Equations by Factoring
32. 33. SOLUTION: SOLUTION: 2
2
2
Factor z from all the three terms.
Use the identity a – b = (a + b)(a – b)
2
Factor 8x – 4x – 12.
Therefore,
Here, a = 8, b = –4 and c = –12.
ac = 8(–12) = –96
34. Find two factors of –96 whose sum is –4.
SOLUTION: –12(8) = –96 and –12 + 8 = –4
2
The GCF of the three terms is 6y . Factor the GCF.
Write –4x as –12x + 8x.
Factor 4x from the first two terms and 4 from the
last two terms.
35. Factor 2x – 3 from the two terms.
SOLUTION: Factor 3 from all the three terms.
2
Factor 5x – 28x – 12.
Here, a = 5, b = –28 and c = –12.
ac = 5(–12) = –60
Therefore,
Find two factors of –60 whose sum is –28.
–30(2) = –60 and –30 + 2 = –28
Write –28x as –30x + 2x.
33. eSolutions Manual - Powered by Cognero
SOLUTION: 2
2
Use the identity a – b = (a + b)(a – b)
Page 5
Factor 5x from the first two terms and 2 from the
last two terms.
Therefore,
4-3 Solving Quadratic Equations by Factoring
35. 36. SOLUTION: Factor 3 from all the three terms.
SOLUTION: Here, a = 12, b = 13 and c = –14.
ac = 12(–14) = –168
2
Factor 5x – 28x – 12.
Find two factors of –168 whose sum is 13.
Here, a = 5, b = –28 and c = –12.
–8(21) = –168 and –8 + 21 = 13
ac = 5(–12) = –60
Write 13x as –8x + 21x.
Find two factors of –60 whose sum is –28.
–30(2) = –60 and –30 + 2 = –28
Factor 4x from the first two terms and 7 from the
last two terms.
Write –28x as –30x + 2x.
Factor 5x from the first two terms and 2 from the
last two terms.
Factor x – 6 from the two terms.
Therefore,
Factor 3x – 2 from the two terms.
Therefore,
37. SOLUTION: Factor out the GCF, 12x.
36. 2
2
Use the identity a – b = (a + b)(a – b) to factor y
– 9.
SOLUTION: Here, a = 12, b = 13 and c = –14.
2
Therefore,
ac = 12(–14) = –168
Find two factors of –168 whose sum is 13.
Solve each equation by factoring.
–8(21) = –168 and –8 + 21 = 13
Write 13x as –8x + 21x.
eSolutions Manual - Powered by Cognero
38. Page 6
SOLUTION: Find the factors of –45 whose sum is 4.
Therefore,
Therefore, the roots are –5 and 9.
4-3 Solving Quadratic Equations by Factoring
Solve each equation by factoring.
39. 38. SOLUTION: Find the factors of –24 whose sum is –5.
SOLUTION: Find the factors of –45 whose sum is 4.
9(–5) = –45 and –5 + 9 = 4
3(–8) = –24 and 3 + (–8) = –5
Write 4x as –5x + 9x.
Write –5x as 3x + (–5x).
Factor x from the first two terms and 9 from the last
two terms.
Factor x from the first two terms and –8 from the
last two terms.
Factor x + 5 from the two terms.
Use the Zero Product Property.
Factor x + 3 from the two terms.
Therefore, the roots are –5 and 9.
Use the Zero Product Property.
39. SOLUTION: Find the factors of –24 whose sum is –5.
Therefore, the roots are –3 and 8.
3(–8) = –24 and 3 + (–8) = –5
Write –5x as 3x + (–5x).
40. SOLUTION: Write the equation with right side equal to zero.
Factor x from the first two terms and –8 from the
last two terms.
2
2
Use the identity a – b = (a + b)(a – b) to factor x
– 121.
2
Use the Zero Product Property.
Factor x + 3 from the two terms.
eSolutions Manual - Powered by Cognero
Page 7
Therefore, the roots are –11 and 11.
and by
8. Factoring
Therefore, the roots are –3
4-3 Solving
Quadratic Equations
Therefore, the roots are –2 and 2.
40. 42. SOLUTION: Write the equation with right side equal to zero.
SOLUTION: Factor out –1.
2
2
Use the identity a – b = (a + b)(a – b) to factor x
– 121.
2
2
Now factor 3x + 10x – 8.
Use the Zero Product Property.
Here, a = 3, b = 10 and c = –8.
ac = 3(–8) = –24
Find two factors of –24 whose sum is 10.
Therefore, the roots are –11 and 11.
12(–2) = –24 and 12 + (–2) = 10
Write 10x as 12x + (–2x).
41. SOLUTION: Write the equation with right side equal to zero.
Factor 3x from the first two terms and –2 from the
last two terms.
2
2
Use the identity a – b = (a + b)(a – b) to factor x
– 4.
2
Factor x + 4 from the two terms.
Use the Zero Product Property.
Use the Zero Product Property.
Therefore, the roots are –2 and 2.
42. Therefore, the roots are
SOLUTION: Factor out –1.
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43. SOLUTION: Factor out –1.
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Therefore, the roots are
4-3 Solving Quadratic Equations by Factoring
43. SOLUTION: Factor out –1.
2
Now factor 8x – 46x + 30.
Here, a = 8, b = –46 and c = 30.
ac = 8(30) = 240
Find two factors of 240 whose sum is –46.
–40(–6) = 240 and –40 + (–6) = –46
Write –46x as –40x + (–6x).
Factor 8x from the first two terms and –6 from the
last two terms.
Factor x – 5 from the two terms.
Use the Zero Product Property.
Therefore, the roots are eSolutions Manual - Powered by Cognero
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4-4 Complex Numbers
CCSS STRUCTURE Simplify.
19. 23. 4i(–6i)
2
SOLUTION: SOLUTION: 11
24. i
SOLUTION: 20. SOLUTION: 25
25. i
21. SOLUTION: SOLUTION: 22. (–3i)(–7i)(2i)
SOLUTION: 26. (10 – 7i) + (6 + 9i)
SOLUTION: 27. (–3 + i) + (–4 – i)
23. 4i(–6i)
2
SOLUTION: SOLUTION: 28. (12 + 5i) – (9 – 2i)
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11
24. i
Page 1
SOLUTION: SOLUTION: 4-4 Complex Numbers
28. (12 + 5i) – (9 – 2i)
33. SOLUTION: SOLUTION: 29. (11 – 8i) – (2 – 8i)
SOLUTION: 30. (1 + 2i)(1 – 2i)
SOLUTION: 34. SOLUTION: 31. (3 + 5i)(5 – 3i)
SOLUTION: 32. (4 – i)(6 – 6i)
SOLUTION: 35. SOLUTION: 33. eSolutions Manual - Powered by Cognero
SOLUTION: Page 2
4-4 Complex Numbers
38. 35. SOLUTION: SOLUTION: 39. SOLUTION: Solve each equation.
36. SOLUTION: 40. SOLUTION: 37. SOLUTION: 41. SOLUTION: 38. eSolutions Manual - Powered by Cognero
SOLUTION: Page 3