9.1 SOLUTIONS
295
CHAPTER NINE
Solutions for Section 9.1
EXERCISES
1. The peanut hits the ground when the height is zero. Since the height is h(t) = −16(t − 8)(t + 8), we know that it is zero
when t = 8 or t = −8. The peanut is dropped when t = 0, so it hits the ground 8 seconds after it is thrown.
2. (a) The coin is thrown at time t = 0, when
Height = h(0) = −16 · 02 + 64 · 0 + 960 = 960 feet.
(b) The coin reaches the ground when the height is zero, which occurs when t = 10 or t = −6. Since the coin is thrown
at t = 0, the coin reaches the ground 10 seconds after it is thrown.
3. (a) If the company has no revenue then
500p(30 − p) = 0.
The expression on the left is zero when one of its factors is zero, that is, when p = 0 and when 30 − p = 0, so
p = 30. The fact that the revenue is zero when p = 0 tells us that if the company charges $0, it makes no money. The
fact that it is zero when p = 30 tells us that if the company charges $30 per item, it makes no sales because no one
will pay that much.
(b) If we write
f (p) = 500p(30 − p) = −500p2 + 15,000p,
it is clear that the coefficient on p2 is negative, so the parabola opens downward. This means that at prices below $0
and above $30, the revenue function is negative. Since it is not reasonable for revenue to be negative from selling a
product, a reasonable domain is 0 ≤ p ≤ 30.
4. If x < −5, then x − 4 and x + 5 are both negative, so their product is positive. If x > 4, then x − 4 and x + 5 are both
positive, so their product is positive. If −5 < x < 4, then x − 4 is negative and x + 5 is positive, so their product is
negative.
5. We first write the function in factored form: g(x) = (x − 8)(x + 7). If x < −7, then x − 8 and x + 7 are both negative,
so their product is positive. If x > 8, then x − 8 and x + 7 are both positive, so their product is positive. If −7 < x < 8,
then x − 8 is negative and x + 7 is positive, so their product is negative.
6. We first write the function in factored form: h(x) = (x − 6)2 . When we square a number, the result is always positive,
unless the number is 0. So h(x) is always positive except when x − 6 = 0, that is x = 6.
7. If x < 1, then x − 1 and x − 2 are both negative, so their product is positive, and k(x) is negative. If x > 2, then x − 1
and x − 2 are both positive, so their product is positive, and k(x) is negative. If 1 < x < 2, then x − 2 is negative and
x − 1 is positive, so their product is negative, and k(x) is positive.
8. The vertex is (h, k), so we look at whether the coordinates of the vertex are positive, negative or zero. The vertex of the
graph is (0, 0), so we see that h = 0 and k = 0.
9. The vertex is (h, k), so we look at whether the coordinates of the vertex are positive, negative or zero. The vertex is on
the negative part of the y-axis, so h = 0 and k < 0.
10. The vertex is (h, k), so we look at whether the coordinates of the vertex are positive, negative or zero. The vertex is on
the positive part of the x-axis, so h > 0 and k = 0.
11. The vertex is (h, k), so we look at whether the coordinates of the vertex are positive, negative or zero. The vertex is in the
quadrant where both x is positive and y is negative, so h > 0 and k < 0.
12. The vertex is (h, k), so we look at whether the coordinates of the vertex are positive, negative or zero. The vertex is in the
quadrant where x is negative and y is negative, so h < 0 and k < 0.
296
Chapter Nine /SOLUTIONS
13. This graph has a vertex at (h, k) = (2, 3) and it opens down so a < 0. This means it matches equation (b).
14. This graph has vertex (h, k) = (2, 3) and it opens up, so it could match (a) or (f). From the graph we see that at
x = 0, y = 7. This agrees with (a) and not with (f), so the matching equation is (a).
15. This graph has vertex (h, k) = (2, 3) and it opens up, so it could match (a) or (f). From the graph we see that at
x = 0, y = 11. This agrees with (f) and not with (a), so the matching equation is (f).
16. The vertex is (h, k) = (3, −2). There is no matching equation.
17. The vertex is (h, k) = (3, 2). This matches (g)
18. The vertex is (h, k) = (−2, 3) and the graph opens down so k < 0. This matches (e).
19. The vertex is (h, k) = (−2, −3) and the graph opens up. This matches (d).
20. The vertex is (h, k) = (−2, 3) and the graph opens up so k > 0. This matches (c).
PROBLEMS
21. (a) The 16 determines the way in which increases in time affect the height. Replacing 16 by a larger number means the
ball is falling faster; replacing it by a smaller number means that the ball is falling slower.
(b) The height of the tower is the ball’s position when t = 0, so it is 100 − 16 · 02 = 100 feet. The height of the tree is
120 − 16 · 02 = 120 feet, so the tree is taller.
(c) The height of the tower is the ball’s position when t = 0, so it is 100 − 16 · 02 = 100 feet. The height of the building
is 100 − 20 · 02 = 100 feet, so the tower and the building are the same height.
Since the coefficient of t2 on the other planet is 20 instead of 16, the height on the other planet decreases faster.
The ball falls faster.
22. (a) The number 8 is the value, w(0), of the function when m = 0, so 8 pounds is the average weight of a newborn.
(b) Since m = 12 when the baby is one year old, the average weight is
w(12) = −0.042m2 + 1.75m + 8 = −0.042(12)2 + 1.75(12) + 8 = 22.952 pounds.
23. (a) The constant a is positive because the parabola faces upward, and c is positive because the y-intercept is on the
positive y-axis.
(b) The vertex is in the third quadrant, where both x and y are negative, so h < 0 and k < 0.
(c) The parabola crosses the x-axis twice, so the expression has two zeros and can be factored. Since both zeros are on
the negative x-axis, r < s < 0.
24. (a) Because the parabola faces downward, a is negative, and c is negative because it is the y-intercept, and the parabola
must cross the y-axis below the x-axis.
(b) Since h and k are the x- and y-coordinates of the vertex, we must have h > 0 and k = 0, since the vertex is on the
positive x-axis.
(c) Since the parabola just touches the x-axis, the equation has one solution, so r = s, and the expression can be factored.
Furthermore, since it touches on the positive side, we have r > 0, so s > 0.
25. (a) Because the parabola faces downward, a is negative, and c is positive because it is the y-intercept, which is positive.
(b) Since h and k are the x- and y-coordinates of the vertex, we must have h = 0 and k > 0, since the vertex is on the
positive y-axis.
(c) The parabola crosses the x-axis twice, so the expression has two zeros and can be factored. Since the zeros are on
opposite sides of the y-axis, r < 0, s > 0.
26. (a) Since the parabola faces upward, a > 0, and since the y-intercept is on the positive y-axis, c > 0.
(b) The vertex is in the second quadrant, where x < 0 and y > 0, so h < 0 and k > 0.
(c) The parabola does not cross the x-axis, so the expression has no real zeros and does not factor.
27. (a) Since the parabola faces upward, we have a > 0, and since it crosses the y-axis at the origin, the y-intercept is 0, so
c = 0.
(b) Since the vertex is in the fourth quadrant, we have h > 0 and k < 0.
(c) The parabolas crosses the axis twice, once at x = 0 and once where x is positive, so the expression can be factored.
Also r and s are not equal, r = 0 and s > 0.
9.2 SOLUTIONS
297
28. (a) Since the parabola faces upward, a > 0, and since the y-intercept is on the positive y-axis, c > 0.
(b) Since h and k are the x- and y-coordinates of the vertex, we must have h < 0 and k = 0, since the vertex is on the
negative x-axis.
(c) Since the parabola just touches the x-axis, the equation has one solution, so r = s, and the expression can be factored.
Furthermore, since it touches on the negative side, we have r < 0, so s < 0.
29. (a) The revenue is the price per widget times the number of widgets. Thus, R(x) = x(1500 − 3x).
(b) It costs the company $5 for each widget. If they sell 1500 − 3x widgets the cost is 5(1500 − 3x). Thus, C(x) =
5(1500 − 3x).
(c) Since P (x) = R(x) − C(x), we substitute the values for R(x) and C(x). Thus, P (x) = x(1500 − 3x) − 5(1500 −
3x). We can factor out (1500 − 3x) to obtain: P (x) = (1500 − 3x)(x − 5).
30. Since the graph is shifted to the right, the value of h must increase. Since the graph is shifted down, the value of k must
decrease.
31. Since the direction that the graph opens changes from upward to downward, the sign of a must change from positive to
negative. Since originally the graph opened up, its vertex must have lain below the x-axis (otherwise the equation would
have had no zeros). After the direction changes to down, the vertex must lie above the x-axis (once again, the equation
would have no zeros otherwise). This means the value of k must increase from a negative value to a positive value.
32. The function g is defined only if the input to the square root operation, (x − 4)(x + 6), is non-negative. The graph of
y = (x − 4)(x + 6) is an upward-opening parabola with x-intercepts (zeros) at x = −6 and x = 4. Thus, the graph lies
on or above the x-axis for x ≤ −6 and x ≥ 4. This means the domain of f is all x such that x ≤ −6 or x ≥ 4.
33. The graph of y = 25+(x−3)2 is an upward-opening parabola with vertex (x, y) = (3, 25).
value of
pThis means the least
√
y on the parabola is y = 25, which occurs at x = 3. Thus, the least value of h is h(3) = 25 + (3 − 3)2 = 25 = 5,
so the range of h is y ≥ 5.
Solutions for Section 9.2
EXERCISES
1. We are shown the vertex of this parabola, so we use the vertex form. The vertex is at the point (6, 15), and the parabola
opens down, so a possible formula is y = −(x − 6)2 + 15.
2. We are shown the x-intercepts of this parabola, so we use the factored form. The x-intercepts are at x = −2, and x = 5
and the parabola opens down, so a possible formula is y = −(x + 2)(x − 5).
3. We are shown the vertex of this parabola, so we use the vertex form. The vertex is at the point (3, −5), and the parabola
opens up, so a possible formula is y = (x − 3)2 − 5.
4. We are shown the x-intercepts of this parabola, so we use the factored form. The x-intercepts are at x = −4 and x = −1,
and the parabola opens up, so a possible formula is y = (x + 4)(x + 1).
5. In vertex form, we have
y = a(x − 2) + 3.
From the y-intercept, we know that at x = 0, y = −4, so
−4 = a(0 − 2)2 + 3
a · 4 = −7
7
a=− ,
4
so
7
y = − (x − 2)2 + 3.
4
298
Chapter Nine /SOLUTIONS
6. From the zeros we know the function has factors (x − 3) and (x + 5), so
y = a(x − 3)(x + 5).
From the y-intercept we know that at x = 0, y = 12, so
12 = a(0 − 3)(0 + 5)
−15a = 12
4
a=− ,
5
so
4
y = − (x − 3)(x + 5).
5
7. In vertex form, we have
y = a(x − 3)2 + 9.
Since the graph passes through the origin, we know that at x = 0, y = 0, so
0 = a(0 − 3)2 + 9
9a = −9
a = −1,
so y = −(x − 3)2 + 9.
8. The x-intercepts of the graph are the same as the function’s zeros, so in factored form we have
y = a(x − 8)(x − 12).
From the y-intercept we know that at x = 0, y = 50, so
50 = a(0 − 8)(0 − 12)
96a = 50
25
50
=
,
a=
96
48
so
y = (25/48)(x − 8)(x − 12).
9. If the length is l feet, the width is l − 6 feet. Since Area = lw, we have Area = l(l − 6) square feet.
10. If the base is x, the height is 2x + 3. Since Area = 1/2bh, we have Area = x(2x + 3)/2 square feet.
11. (a) We multiply out and combine like terms:
(x + 3)2 − 1 = (x2 + 6x + 9) − 1 = x2 + 6x + 8.
The standard form of this expression is x2 + 6x + 8.
We factor the standard form:
x2 + 6x + 8 = (x + 2)(x + 4).
The factored form of this expression is (x + 2)(x + 4).
(b) We substitute x = 0 into each form:
Vertex: (0 + 3)2 − 1 = 32 − 1 = 9 − 1 = 8.
Standard: 02 + 6 · 0 + 8 = 0 + 0 + 8 = 8.
Factored: (0 + 2)(0 + 4) = 2 · 4 = 8.
As expected, since the expressions are all equivalent, we get the same value no matter which form we use. We
substitute x = 3 into each form:
9.2 SOLUTIONS
299
Vertex: (3 + 3)2 − 1 = 62 − 1 = 36 − 1 = 35.
Standard: 32 + 6 · 3 + 8 = 9 + 18 + 8 = 35.
Factored: (3 + 2)(3 + 4) = 5 · 7 = 35.
As expected, since the expressions are all equivalent, we get the same value no matter which form we use.
12. (a) We multiply out and combine like terms:
(x − 2)2 − 25 = (x2 − 4x + 4) − 25 = x2 − 4x − 21.
The standard form of this expression is x2 − 4x − 21.
We factor the standard form:
x2 − 4x − 21 = (x + 3)(x − 7).
The factored form of this expression is (x + 3)(x − 7).
(b) We substitute x = 0 into each form:
Vertex: (0 − 2)2 − 25 = (−2)2 − 25 = 4 − 25 = −21.
Standard: 02 − 4 · 0 − 21 = 0 − 0 − 21 = −21.
Factored: (0 + 3)(0 − 7) = 3 · (−7) = −21.
As expected, since the expressions are all equivalent, we get the same value no matter which form we use. We
substitute x = 3 into each form:
Vertex: (3 − 2)2 − 25 = 12 − 25 = 1 − 25 = −24.
Standard: 32 − 4 · 3 − 21 = 9 − 12 − 21 = −24.
Factored: (3 + 3)(3 − 7) = 6 · (−4) = −24.
As expected, since the expressions are all equivalent, we get the same value no matter which form we use.
13. Since f (x) = x(x − 3) = x2 − 3x, we have a = 1, b = −3 and c = 0.
14. Since q(m) = (m − 7)2 = m2 − 14m + 49, we have a = 1, b = −14 and c = 49.
15. We multiply out:
f (n) = (n − 4)(n + 7) = n2 − 4n + 7n − 28 = n2 + 3n − 28.
We have a = 1, b = 3 and c = −28.
√ 2
√
√
2p = − 2p2 + 0p + 1, we have a = − 2, b = 0 and c = 1.
16. Since g(p) = 1 −
17. We multiply out and combine like terms:
m(t) = 2(t − 1)2 + 12 = 2(t2 − 2t + 1) + 12
= 2t2 − 4t + 2 + 12
= 2t2 − 4t + 14.
We have a = 2, b = −4 and c = 14.
18. We multiply out and combine like terms:
p(q) = (q + 2)(3q − 4) = 3q 2 + 6q − 4q − 8
= 3q 2 + 2q − 8.
We have a = 3, b = 2 and c = −8.
19. Since
h(x) = (x − r)(x − s) = x2 − (r + s)x + rs,
we have a = 1, b = −(r + s) and c = rs.
300
Chapter Nine /SOLUTIONS
20. Since
p(x) = a(x − h)2 + k = a(x2 − 2hx + h2 ) + k = ax2 − 2ahx + ah2 + k,
we have a = a, b = −2ah and c = ah2 + k.
21. We multiply out and then combine like terms:
q(p) = (p − 1)(p − 6) + p(3p + 2) = (p2 − p − 6p + 6) + (3p2 + 2p)
= 4p2 − 5p + 6.
We have a = 4, b = −5 and c = 6.
22. We multiply out and combine like terms:
h(t) = 3(2t − 1)(t + 5) = 3(2t2 − t + 10t − 5)
= 3(2t2 + 9t − 5)
= 6t2 + 27t − 15.
We have a = 6, b = 27 and c = −15.
23. Since 3 · 5 = 15 and 3 + 5 = 8, we have x2 + 8x + 15 = (x + 3)(x + 5) so the factored form is (x + 3)(x + 5).
24. Since −4 · 6 = −24 and −4 + 6 = 2, we have x2 + 2x − 24 = (x − 4)(x + 6) so the factored form is (x − 4)(x + 6).
25. We first multiply out and combine like terms:
(x − 4)2 − 4 = (x2 − 8x + 16) − 4 = x2 − 8x + 12.
We factor x2 − 8x + 12. Since (−2) · (−6) = 12 and −2 + −6 = −8, we have x2 − 8x + 12 = (x − 2)(x − 6). The
original expression is equivalent to the factored form (x − 2)(x − 6).
Alternatively, we could recognize (x − 4)2 − 4 as the difference between two squares:
(x − 4)2 − 4 = (x − 4)2 − 22
= ((x − 4) − 2)((x − 4) + 2)
= (x − 6)(x − 2).
26. We have
x2 + 7x + 10 = (x + 2)(x + 5)
27. This is the difference of two squares:
4z 2 − 49 = (2z − 7)(2z + 7)
28. This is a perfect square:
9t2 + 60t + 100 = (3t + 10)(3t + 10) = (3t + 10)2 .
29. We first multiply out and combine like terms:
(x + 2)(x − 3) + 2(x − 3) = (x2 − x − 6) + (2x − 6)
= x2 + x − 12.
We factor x2 + x − 12. Since −3 · 4 = −12 and −3 + 4 = 1, we have x2 + x − 12 = (x − 3)(x + 4). The original
expression is equivalent to the factored form (x − 3)(x + 4).
Alternatively, we could recognize that (x − 3) is a common factor, and find
(x + 2)(x − 3) + 2(x − 3) = (x − 3)((x + 2) + 2) = (x − 3)(x + 4).
9.2 SOLUTIONS
301
30. We have a = 6, b = 31, c = 40. The product ab = 6 · 40 = 240. We need factors of 240 that add to 31. We see that
16(15) = 240 and 16 + 15 = 31, so
6w2 + 31w + 40 = 6w2 + 15w + 16w + 40
= 3w(2w + 5) + 8(2w + 5)
= (3w + 8)(2w + 5).
31. This is already in vertex form with a = −2, h = 5, k = 5.
32. Writing this as
y=
1
(x − (−4))2 + (−7),
5
we have a = 1/5, h = −4, k = −7.
33. Writing this as
y = (2(x − 2))2 + 6
= 22 (x − 2)2 + 6
= 4(x − 2)2 + 6,
so a = 4, h = 2, k = 6.
34. Writing this as
y − 5 = (2 − x)2
y = (−1(x − 2))2 + 5
= (−1)2 (x − 2)2 + 5
= (x − 2)2 + 5,
we have a = 1, h = 2, k = 5.
35. We first subtract 20 from both sides:
y − 20 = x2 + 12x.
Next, we complete the square by adding (12/2)2 = 36 to both sides:
y − 20 + 36 = x2 + 12x + 36
y + 16 = (x + 6)2
y = (x − (−6))2 + (−16).
We have a = 1, h = −6, k = −16.
36. We first subtract 17 from both sides, then divide both sides by 4:
y − 17
= x2 + 6x.
4
We now complete the square by adding (6/2)2 = 9 to both sides:
y − 17
+9
4
y − 17
+9
4
y − 17
4
y − 17
= x2 + 6x + 9
= (x + 3)2
= (x + 3)2 − 9
= 4(x + 3)2 − 36
y = 4(x + 3)2 − 19
= 4(x − (−3))2 + (−19),
so a = 4, h = −3, k = −19.
302
Chapter Nine /SOLUTIONS
37. We have
y − 5 = x2 + 2x + 1
y − 5 = (x + 1)2
y = (x + 1)2 + 5,
so a = 1, h = −1, k = 5.
38. We have
y − 12 = 2x2 + 8x + 8
y − 12 = 2(x2 + 4x + 4)
y − 12 = 2(x + 2)2
y = 2(x + 2)2 + 12,
so a = 2, h = −2, k = 12.
39. We have
y = x2 + 6x + 4
y + 5 = x2 + 6x + 4 + 5
y + 5 = (x + 3)2
y = (x + 3)2 − 5,
so a = 1, h = −3, k = −5.
40. We have
y = 2x2 + 20x + 12
= 2(x2 + 10x) + 12
= 2(x2 + 10x + 25) − 50 + 12
= 2(x + 5)2 − 38,
so a = 2, h = −5, k = −38.
41. We have
y + 2 = x2 − 3x
2
3
3
= x2 − 3x +
y+2+
2
2
3 2
9
y+2+ = x−
4
2
3 2 17
y = x−
,
−
2
4
2
so a = 1, h = 3/2, k = −17/4.
42. We have
y = 3x2 + 7x + 5
7
y − 5 = 3 x2 + x
3
2
7
7
7
y − 5 = 3 x 2 + x + ( )2 − 3
3
6
6
7 2 49
−
y−5 = 3 x+
6
12
2
7
11
y = 3 x+
,
+
6
12
so a = 3, h = −7/6, k = 11/12.
9.2 SOLUTIONS
303
43. We see that the y-intercept is 12, since the equation is in standard form. To find the x-intercepts, we factor:
y = 3x2 + 15x + 12
y = 3(x2 + 5x + 4)
y = 3(x + 4)(x + 1).
Thus, the x-intercepts are at −4 and −1.
44. We see that the y-intercept is −10, since the equation is in standard form. To find the x-intercepts, we factor:
y = x2 − 3x − 10
y = (x − 5)(x + 2).
Thus, the x-intercepts are at 5 and −2.
45. We see that the y-intercept is 2, since the equation is in standard form. To find the x-intercepts, we complete the square:
0 = x2 + 5x + 2
−2 = x2 + 5x
25
25
= x2 + 5x +
−2 +
4
4
5 2
17
= x+
4
2
Thus, the x-intercepts are at ±
5
x+ = ±
2
r
17
4
x=±
r
5
17
− .
4
2
p
17/4 − 5/2.
46. We see that the y-intercept is −30, since the equation is in standard form. To find the x-intercepts, we complete the square:
0 = 3x2 + 18x − 30
0 = 3(x2 + 6x − 10)
10 = x2 + 6x
10 + 9 = x2 + 6x + 9
√
Thus, the x-intercepts are at ± 19 − 3.
19 = (x + 3)2
√
x + 3 = ± 19
√
x = ± 19 − 3.
47. Put the expression on the right-hand side in vertex form by completing the square:
x(x − 1) = x2 − x = (x − 1/2)2 − 1/4.
So the vertex is at x = 1/2, y = −1/4.
48. Put the expression on the right-hand side in vertex form by completing the square:
x2 − 2x + 3 = (x − 1)2 − 1 + 3 = (x − 1)2 + 2.
So the vertex is at x = 1, y = 2.
49. Reordering the terms gives y = −x2 + 2, which is already in vertex form. The vertex is at x = 0, y = 2.
304
Chapter Nine /SOLUTIONS
50. Put the expression on the right-hand side in vertex form by completing the square:
2x + 2 − x2 = −(x2 − 2x) + 2 = −((x − 1)2 − 1) + 2 = −(x − 1)2 + 1 + 2 = −(x − 1)2 + 3.
So the vertex is at x = 1, y = 3.
51. This function is in vertex form, and the vertex is (h, k) = (3, 10). Since a is positive, the minimum value of the function
is k = 10.
52. We must complete the square to put the function in vertex form.
g(x) = x2 − 2x − 8
= x2 − 2x + 1 − 1 − 8
= (x − 1)2 − 9.
The vertex is (h, k) = (1, −9). Since a is positive, the minimum value of the function is k = −9.
53. We must first expand, and then complete the square to put the function in vertex form.
h(x) = (x − 5)(x − 1)
= x2 − 6x + 5
= x2 − 6x + 32 − 32 + 5
= (x − 3)2 − 4.
The vertex is (h, k) = (3, −4). Since a is positive, the minimum value of the function is k = −4.
54. This function is in vertex form, and the vertex is (h, k) = (4, 7). Since a is negative, the vertex is a maximum. There is
no minimum value of this function.
PROBLEMS
55. (a) We show that the factored and vertex forms are equivalent to the standard form. For the factored form:
−2(p − 3)(p − 9) = −2(p2 − 3p − 9p + 27)
= −2(p2 − 12p + 27)
= −2p2 + 24p − 54.
For the vertex form:
−2(p − 6)2 + 18 = −2(p2 − 12p + 36) + 18
= −2p2 + 24p − 72 + 18
= −2p2 + 24p − 54.
In each case, we get the standard form given, so all three expressions are equivalent.
(b) The factored form gives the values of p that make the profit zero. Since factored form is −2(p − 3)(p − 9), the profit
is zero when p = 3 or p = 9. The company breaks even if the price charged for the product is $3 or $9.
(c) The standard form is the easiest one to use to find the profit when the price is zero. Substituting p = 0 into the
standard form −2p2 + 24p − 54, we see that the profit is −54 (in thousands of dollars) when the price is zero. If the
company gives the product away for free, it loses $54,000.
(d) The vertex form shows us what price maximizes profit. From the expression −2(p − 6)2 + 18, we see that the
maximum profit is 18 thousand dollars, and it occurs when p = 6. The company should charge a price of $6 for this
product.
56. A quadratic function whose zeros are 3 and 4 is f (x) = (x − 3)(x − 4). We multiply out and combine like terms to write
this function in standard form:
f (x) = (x − 3)(x − 4) = x2 − 3x − 4x + 12 = x2 − 7x + 12.
Other possibilities are obtained by multiplying this by any non-zero constant.
9.2 SOLUTIONS
305
57. We want the quadratic function to be zero when x = a + 1 or x = 3a, so f (x) = (x − (a + 1))(x − 3a) is one possibility.
(Other possibilities are obtained by multiplying this by any non-zero constant.) Thus,
f (x) = (x − (a + 1))(x − 3a) = x2 − 3ax − (a + 1)x + 3a(a + 1) = x2 − (4a + 1)x + 3a(a + 1).
58. We see that (a), (c), and (d) are equivalent:
(t + 3)2 = (t + 3)(t + 3)
= t2 + 6t + 9
so (a) and (c) are equivalent
2
t(t + 9) − 3(t − 3) = t + 9t − 3t + 9
= t2 + 6t + 9
so (d) and (c) are equivalent
Likewise, (b) and (f) are equivalent:
2(t2 + 9)
2t2 + 18
=
2
2
= t2 + 9
so (f) and (b) are equivalent
Finally, expression (e) is not equivalent to any of the others.
59. (a) In standard form, we have
y = x2 − 3x − 7x + 21
= x2 − 10x + 21,
so a = 1, b = −10, c = 21.
(b) This is already almost in factored form, since there is a common factor of (x − 3). We have:
y = (x − 3)(x − 7),
so a = 1, r = 3, s = 7.
(c) To place this in vertex form, we begin with standard form, which from part (a) is y = x2 − 10x + 21:
y − 21 = x2 − 10x
y − 21 + 25 = x2 − 10x + 25
because (−10/2)2 = 25
y = (x − 5)2 + (−4),
so a = 1, h = 5, k = −4.
60. (a) This is already almost in standard form. Writing
y = 6x2 − 23x + 21,
we have a = 6, b = −23, c = 21.
(b) We need numbers whose product is ac = 6(21) = 126 and whose sum is b = −23. We find that −9(−14) = 126
and −9 + (−14) = −23, so we write
y = 6x2 − 9x − 14x + 21
= 3x(2x − 3) − 7(2x − 3)
= (3x − 7)(2x − 3)
Although this is factored, it is not quite in factored form. We have
7
3
·2 x−
3
2
3
7
x−
,
= 6 x−
3
2
y = 3 x−
so a = 6, r = 7/3, s = 3/2
306
Chapter Nine /SOLUTIONS
(c) We complete the square:
= 6x2 − 23x
23
= x2 −
·x
6
2
23
23
2
=x −
·x+
6
12
23 2 529
−
= x−
12 2 144 529
23
−
y = 6 x−
+ 21
12
144
2
504
529
23
+
−
= 6 x−
12
24
25 24
23 2
= 6 x−
+ −
,
12
24
y − 21
y − 21
6 2
y − 21
23
+
6
12
y − 21
6
because
1
2
·
23
6
=
23
12
so a = 6, h = 23/12, k = −25/24.
61. The vertex form F (x) = a(x − h)2 + k allows us to easily see the largest or smallest value. We use k = 100. Since we
want all other values of the expression to be less than 100, the expression a(x − h)2 must be negative or zero, so a must
be negative. Finally, since the largest value should occur at x = 3, we take h = 3. The function F (x) = −(x − 3)2 + 100
takes its largest value of 100 at x = 3, and is one possibility. Other possibilities are of the form F (x) = a(x − 3)2 + 100,
where a < 0.
We multiply out and combine like terms to write this function in standard form:
F (x) = −(x − 3)2 + 100 = −(x2 − 6x + 9) + 100
= −x2 + 6x − 9 + 100
= −x2 + 6x + 91.
62. (a) She sells no chairs when 1200 − 3p = 0, that is, when p = 400. So she prices herself out of the market if she charges
$400 for a chair.
(b) Her revenue is
Price × Quantity = p(1200 − 3p)
= −3p2 + 1200p
= −3(p2 − 400p)
= −3(p − 200)2 + 120,000
completing the square.
So her maximum revenue occurs when p = 200. So she should charge $200 to maximize the revenue.
63. (a) If we let w represent the width of the rectangle, the length is 2w. Since the walk is 2 feet wide on each side, the width
of the pool and the walk is w + 4. The length of the pool and the walk is 2w + 4. We substitute this information into
the formula for area and solve for w:
A = lw
1056 = (2w + 4)(w + 4)
1056 = 2w2 + 12w + 16
1040 = 2w2 + 12w
520 = w2 + 6w
520 + 9 = w2 + 6w + 9
529 = (w + 3)2
√
w + 3 = ± 529
w = −3 ± 23
w = 20 or
w = −26.
9.2 SOLUTIONS
307
Since we are dealing with dimensions of a rectangle, we are only interested in the positive answer. Thus, the dimensions of the pool are 20 feet by 40 feet.
(b) The area of the pool and the walk is 1056 square feet. The area of the pool is 20 · 40 = 800 square feet. Thus, the area
of the walk is 1056 − 800 = 256 square feet. At a cost of $10 per square foot, the walk would cost 256 · 10 = 2560
dollars.
64. The factored form of this function is y = (x + 2)(x + 6). The x-intercepts of this parabola are at x = −2 and x = −6.
Since the coefficient of x2 is positive, the parabola opens up. A possible graph is shown in Figure 9.1.
y
x
−6
−2
Figure 9.1
65. The factored form of this function is y = (x + 1)(x − 7). The x-intercepts of this parabola are at x = −1 and x = 7.
Since the coefficient of x2 is positive, the parabola opens up. A possible graph is shown in Figure 9.2.
y
7
−1
x
Figure 9.2
66. Taking the product of a = 8 and c = −15 gives ac = −120. We need factors of −120 that add to b = −2. We see that
−12(10) = −120 and −12 + 10 = −2, so we write
y = 8x2 + 10x − 12x − 15
= 2x(4x + 5) − 3(4x + 5)
= (2x
− 3)(4x
+5) 5 3
= 2 x−
·4 x+
2
4 3
5
x− −
,
= 8 x−
2
4
so a = 8, r = 3/2, s = −5/4.
67. We first isolate the x2 term:
y = 8x2 − 2x − 15
y + 15 = 8x2 − 2x
1
y + 15
= x2 − · x.
8
4
308
Chapter Nine /SOLUTIONS
Taking half the coefficient of x, we complete the square:
1
1 2
1 2
y + 15
= x2 − · x + −
+ −
8
8
4
8
y + 15
1
1 2
+
= x−
8
64
8
1
1 2
y + 15
−
= x−
8
8
64
1 2 1
−
y + 15 = 8 x −
8
8
1 2
1
y = 8 x−
− 15 −
8
8
121
1 2
+ −
y = 8 x−
8
8
so a = 8, h = 1/8, k = −121/8.
68. We have
y = 8x2 − 2x − 15
= (8x2 − 2x) + (−15)
grouping terms involving x
= −2x(−4x + 1) + (−15), factoring −2x to leave 1
so k = −2, v = −4, w = −15.
69. The first term contributes 2x2 , the second term contributes −5x2 , and the last two terms are linear, so do not contribute
anything. Since 2x2 − 5x2 = −3x2 , the coefficient of x2 is −3.
70. We start with x and perform the operations in order:
Add 5: x + 5
Multiply by x: x(x + 5)
Subtract 2: x(x + 5) − 2.
We put the result in standard form:
x(x + 5) − 2 = x2 + 5x − 2.
71. We start with x and perform the operations in order:
Subtract 3: x − 3
Multiply by x: x(x − 3)
Add 2: x(x − 3) + 2
Multiply by 5: 5(x(x − 3) + 2).
We put the result in standard form:
5(x(x − 3) + 2) = 5(x2 − 3x + 2) = 5x2 − 15x + 10.
9.3 SOLUTIONS
309
Solutions for Section 9.3
EXERCISES
1. If we find the radius of the garden, we can then find the circumference, so we first solve the quadratic equation πr 2 = 80
for the radius of the garden:
πr 2 = 80
80
r2 =
π
r=±
r
80
= ±5.046 meters.
π
Since the radius cannot be negative, we see that it must be 5.046 meters. We multiply the radius by 2π to get the circumference of the garden:
r
80
= 31.707 meters.
2π
π
So you need 31.707 meters of fence.
2.
2x2 − 0.3x = 9
2x2 − 0.3x − 9 = 0,
3.
so a = 2, b = −0.3, c = −9.
3x − 2x2 = −7
−2x2 + 3x + 7 = 0,
so a = −2, b = 3, c = 7.
4. Writing this as −x2 + 0 · x + 4 = 0, we see that a = −1, b = 0, c = 4.
5.
A=π
x2
π·
−A = 0
4
x 2
2
π 2
x + 0 · x − A = 0,
4
6.
so a = π/4, b = 0, c = −A.
−2(2x − 3)(x − 1) = 0
−2(2x2 − 2x − 3x + 3) = 0
−2(2x2 − 5x + 3) = 0
−4x2 + 10x − 6 = 0,
so a = −4, b = 10, c = −6.
310
Chapter Nine /SOLUTIONS
7.
1
(1 − x)
1−x
4 − 3x
2−x
4 − 3x
= (1 − x) ·
2−x
(1 − x)(4 − 3x)
=
2−x
4 − 7x + 3x2
=
2−x
4 − 7x + 3x2
= (2 − x) ·
2−x
= 4 − 7x + 3x2
−4 =
1
−4
1−x
1 − 4(1 − x)
4x − 3
(2 − x)(4x − 3)
8x − 6 − 4x2 + 3x
7x2 − 18x + 10 = 0,
8.
so a = 7, b = −18, c = 10.
7x(7 − x − 5(x − 7)) = (2x − 3)(3x − 2)
7x(7 − x − 5x + 35) = 6x2 − 4x − 9x + 6
7x(−6x + 42) = 6x2 − 13x + 6
−42x2 + 294x = 6x2 − 13x + 6
48x2 − 307x + 6 = 0,
9.
so a = 48, b = −307, c = 6.
t2 x − x2 t3 + tx2 − t3 − 4x2 − 3x = 5
tx2 − x2 t3 − 4x2 + t2 x − 3x − t3 − 5 = 0
(t − t3 − 4)x2 + (t2 − 3)x − t3 − 5 = 0.
10.
group like terms
factor
So a = t − t3 − 4, b = t2 − 3, c = −t3 − 5.
5x (x2 + 2x + 1) − 2 = 5x (x2 + 2x + 1) − x
5x(x2 + 2x − 1) = 5x(x2 + x + 1)
5x3 + 10x2 − 5x = 5x3 + 5x2 + 5x
5x2 − 10x = 0,
so a = 5, b = −10, c = 0.
11. We take the square root of each side, getting
√
x = ± 9 = ±3.
12. We rewrite the equation as
x2 = 7,
and take the square root of each side, getting
√
x = ± 7 = ±2.648.
9.3 SOLUTIONS
311
13. We subtract 3 from both sides
x2 = 14,
and take the square root of each side, getting
√
x = ± 14 = ±3.742.
14. We add 4 to each side:
(x + 2)2 = 4.
We take the square root of each side, getting
√
x + 2 = ± 4 = ±2,
and subtract 2 from both sides:
x + 2 = ±2
x = ±2 − 2
x = 0 and − 4.
15. We add 6 to each side:
We take the square root of each side, getting
(x − 3)2 = 16.
√
x − 3 = ± 16 = ±4.
We add 3 to both sides:
x − 3 = ±4
x = ±4 + 3
x = 7 and − 1.
16. To solve this equation, we subtract 5 from both sides to obtain (x − 1)2 = −5. Since a square is always positive, this
equation has no solution.
17.
(x − 5)2 = 6
√
x−5 = ± 6
√
x = 5 ± 6.
18.
7(x − 3)2 = 21
(x − 3)2 = 3
√
x−3 = ± 3
√
x = 3 ± 3.
312
Chapter Nine /SOLUTIONS
19.
1 − 4(9 − x)2 = 13
−4(9 − x)2 = 12
(9 − x)2 = −3.
This has no solutions.
20.
2(x − 1)2 = 5
5
(x − 1)2 =
2
x−1 = ±
r
x = 1±
5
2
r
5
.
2
21. This gives two equations:
2
(x − 3)2 + 1
= 16
√
(x − 3) + 1 = ± 16 = ±4
2
(x − 3)2 = −1 ± 4.
The first equation, (x − 3)2 = −1 + 4, gives
(x − 3)2 = 3
√
x−3 = ± 3
√
x = 3 ± 3.
The second equation, (x − 3)2 = −1 − 4, gives
(x − 3)2 = −5,
and has no solutions.
22.
2
x2 − 5
−5 = 0
2
x2 − 5
=5
√
x −5 = ± 5
√
x2 = 5 ± 5
2
x=±
Note that this represents 4 different solutions:
p
√
x=p
− 5− 5
√
5− 5
x= p
√
x=p
− 5+ 5
√
x = 5+ 5
p
5±
√
5.
= −1.663
= 1.663
= −2.690
= 2.690.
9.3 SOLUTIONS
23. The square has already been completed on the left-and side, so we can write
x2 + 6x + 9 = 4
(x + 3)2 = 4
x = −3 ±
√
4
= −3 ± 2,
so x = −5, −1.
24. Completing the square, we have
x2 − 12x − 5 = 0
x2 − 12x = 5
x2 − 12x + 62 = 5 + 36
(x − 6)2 = 41
x = 6±
√
41.
25. Completing the square, we have
2x2 + 3x − 1
3
x2 + x
2
3 2
3
2
x + x+
2
4
3 2
x+
4
3
x+
4
=0
1
=
2
9
1
= +
2
16
17
=
16
√
17
=±
4
√
−3 ± 17
.
x=
4
26. Using the quadratic formula with a = −2, b = 5, c = −5 gives
x=
=
−5 ±
p
52 − 4(−2)(−5)
2(−2)
√
−5 ± −15
,
2(−2)
so there are no solutions.
27. Using the quadratic formula with a = 1, b = 5, c = −7 gives
x=
=
−5 ±
p
52 − 4(1)(−7)
2(1)
√
−5 ± 53
.
2
28.
2
(2x + 5)(x − 3) = 7
2x − 6x + 5x − 15 − 7 = 0
2x2 − x − 22 = 0,
313
314
Chapter Nine /SOLUTIONS
so by the quadratic formula
x=
=
−(−1) ±
1±
√
p
177
4
(−1)2 − 4(2)(−22)
2(2)
.
29. (a) Solve in steps:
Step 1. Add −12 to both sides, giving x2 + 8x = −12.
Step 2. Since 8/2 = 4 and 42 = 16, we add 16 to both sides:
x2 + 8x + 16 = −12 + 16.
Step 3. Rewriting the left side and combining terms on the right side gives
(x + 4)2 = 4.
Step 4. Taking square roots, we get
(x + 4) =
√
4=2
and
√
(x + 4) = − 4 = −2,
so
x = −4 + 2 = −2
and
(b) We have a = 1, b = 8, c = 12, so
x=
−8 ±
x = −4 − 2 = −6.
√
82 − 4 · 1(12)
−8 ± 16
−8 ± 4
=
=
.
2·1
2
2
p
Thus, x = −2 and x = −6, which agrees with part (a).
30. (a) Solve in steps:
Step 1. Add 15 to both sides, giving x2 − 10x = 15.
Step 2. Since −10/2 = −5 and (−5)2 = 25, we add 25 to both sides:
x2 − 10x + 25 = 15 + 25.
Step 3. Rewriting the left side and combining terms on the right side gives
(x − 5)2 = 40.
Step 4. Taking square roots, we get
(x − 5) =
√
so
x=5+
(b) We have a = 1, b = −10, c = −15, so
x=
But,
−(−10) ±
40
√
40
and
and
√
(x − 5) = − 40,
x=5−
√
40.
√
√
(−10)2 − 4 · 1(−15)
160
10 ± 160
=
= 5±
.
2·1
2
2
p
√
√
√
160
4 × 40
=
= 40,
2
2
√
√
so, x = 5 + 40 and x = 5 − 40, which agree with part (a).
9.3 SOLUTIONS
31. (a) We first divide both sides by 2, giving x2 + 8x − 12 = 0.
Step 1. Add 12 to both sides, giving x2 + 8x = 12.
Step 2. Since 8/2 = 4 and 42 = 16, we add 16 to both sides:
x2 + 8x + 16 = 12 + 16.
Step 3. Rewriting the left side and combining terms on the right side gives
(x + 4)2 = 28.
Step 4. Taking square roots, we get
(x + 4) =
√
so
x = −4 +
(b) We have a = 2, b = 16, c = −24, so
x=
−16 ±
28
√
and
28
√
(x + 4) = − 28,
x = −4 −
and
√
28.
√
√
(16)2 − 4 · 2(−24)
448
−16 ± 448
=
= −4 ±
.
2·2
4
4
p
But,
√
√
√
448
16 × 28
=
= 28,
4
4
√
√
so, x = −4 + 28 and x = −4 − 28, which agrees with part (a).
32. (a) Solve in steps:
Step 1. Add −5 to both sides, giving x2 + 7x = −5.
Step 2. Since (7/2)2 = 49/4, we add 49/4 to both sides:
x2 + 7x +
49
49
= −5 +
.
4
4
Step 3. Rewriting the left side and combining terms on the right side gives
7
2
29
4
and
x+
2
29
.
4
=
Step 4. Taking square roots, we get
x+
7
2
=
r
so
7
x=− +
2
√
29
2
and
x+
7
2
=−
7
x=− −
2
r
29
,
4
√
29
.
2
(b) We have a = 1, b = 7, c = 5, so
x=
which agrees with part (a).
−7 ±
√
√
(7)2 − 4 · 1(5)
−7 ± 29
7
29
=
=− ±
,
2·1
2
2
2
p
315
316
Chapter Nine /SOLUTIONS
33. (a) Solve in steps:
Step 1. Add −2 to both sides, giving x2 − 9x = −2.
Step 2. Since (−9/2)2 = 81/4, we add 81/4 to both sides:
x2 − 9x +
81
81
= −2 +
.
4
4
Step 3. Rewriting the left side and combining terms on the right side gives
2
9
2
73
4
and
x−
73
.
4
=
Step 4. Taking square roots, we get
9
2
x−
=
r
so
x=
(b) We have a = 1, b = −9, c = 2, so
x=
−(−9) ±
9
+
2
√
73
2
x−
x=
and
9
2
9
−
2
=−
r
73
,
4
√
73
.
2
√
√
(−9)2 − 4 · 1(2)
9 ± 73
9
73
=
= ±
,
2·1
2
2
2
p
which agrees with part (a).
34. (a) Solve in steps:
Step 1. Add 8 to both sides, giving x2 + 17x = 8.
Step 2. Since (17/2)2 = 289/4, we add 289/4 to both sides:
x2 + 17x +
289
289
=8+
.
4
4
Step 3. Rewriting the left side and combining terms on the right side gives
x+
17
2
2
321
.
4
=
Step 4. Taking square roots, we get
17
x+
2
=
r
so
17
+
x=−
2
(b) We have a = 1, b = 17, c = −8, so
x=
which agrees with part (a).
−17 ±
321
4
√
321
2
and
and
17
x+
2
=−
17
x=−
−
2
r
321
,
4
√
321
.
2
√
√
(17)2 − 4 · 1(−8)
−17 ± 321
−17
321
=
=
±
,
2·1
2
2
2
p
35. (a) Solve in steps:
Step 1. Add −10 to both sides, giving x2 − 22x = −10.
Step 2. Since −22/2 = −11 and (−11)2 = 121, we add 121 to both sides:
x2 − 22x + 121 = −10 + 121.
Step 3. Rewriting the left side and combining terms on the right side gives
(x − 11)2 = 111.
9.3 SOLUTIONS
Step 4. Taking square roots, we get
(x − 11) =
√
so
x = 11 +
111
√
111
and
and
(b) We have a = 1, b = −22, c = 10, so
x=
−(−22) ±
√
(x − 11) = − 111,
x = 11 −
√
111.
√
√
√
(−22)2 − 4 · 1(10)
22 ± 444
4 · 111
=
= 11 ±
= 11 ± 111,
2·1
2
2
p
which agrees with part (a).
36. (a) We first divide both sides by 2, giving x2 − 16x + 7/2 = 0.
Step 1. Add −7/2 to both sides, giving x2 − 16x = −7/2.
Step 2. Since −16/2 = −8 and (−8)2 = 64, we add 64 to both sides:
x2 − 16x + 64 = −
7
+ 64.
2
Step 3. Rewriting the left side and combining terms on the right side gives
(x − 8)2 =
121
.
2
Step 4. Taking square roots, we get
(x − 8) =
r
so
x=8+
121
2
r
121
2
and
and
(x − 8) = −
x=8−
r
r
121
,
2
121
.
2
(b) We have a = 2, b = −32, c = 7, so
x=
−(−32) ±
But
√
√
√
(−32)2 − 4 · 2(7)
968
968
32 ± 968
32
=
=
±
=8±
.
2·2
4
4
4
4
p
√
r
r
√
968
968
968
121
=
= √
=
,
4
16
2
16
which agrees with part (a).
37. (a) We first divide both sides by 3, giving x2 + 6x + 2/3 = 0.
Step 1. Add −2/3 to both sides, giving x2 + 6x = −2/3.
Step 2. Since 6/2 = 3 and 32 = 9, we add 9 to both sides:
x2 + 6x + 9 = −
2
+ 9.
3
Step 3. Rewriting the left side and combining terms on the right side gives
(x + 3)2 =
25
.
3
Step 4. Taking square roots, we get
(x + 3) =
r
so
x = −3 +
25
3
r
25
3
and
and
(x + 3) = −
x = −3 −
r
r
25
,
3
25
.
3
317
318
Chapter Nine /SOLUTIONS
(b) We have a = 3, b = 18, c = 2, so
x=
−18 ±
But
√
√
√
(18)2 − 4 · 3(2)
−18 ± 300
−18
300
300
=
=
±
= −3 ±
.
2·3
6
6
6
6
p
√
r
√
300
300
25
=
= √
,
6
3
36
which agrees with part (a).
38. (a) We first divide both sides by 5, giving x2 + (17/5)x + 1/5 = 0.
Step 1. Add −1/5 to both sides, giving x2 + (17/5)x = −1/5.
Step 2. Since (17/5)/2 = 17/10 and (17/10)2 = 289/100, we add 289/100 to both sides:
289
1
289
17
x+
=− +
.
5
100
5
100
x2 +
Step 3. Rewriting the left side and combining terms on the right side gives
x+
17
10
2
269
.
100
=
Step 4. Taking square roots, we get
17
x+
10
=
r
so
x=−
17
+
10
269
100
√
269
10
and
17
x+
10
x=−
and
=−
17
−
10
r
269
,
100
√
269
.
10
(b) We have a = 5, b = 17, c = 1, so
x=
−17 ±
√
√
(17)2 − 4 · 5(1)
−17 ± 269
−17
269
=
=
±
,
2·5
10
10
10
p
which agrees with part (a).
39. (a) We first divide both sides by 6, giving x2 + (11/6)x − 10/6 = 0.
Step 1. Add 10/6 to both sides, giving x2 + (11/6)x = 5/3.
Step 2. Since (11/6)/2 = 11/12 and (11/12)2 = 121/144, we add 121/144 to both sides:
x2 +
11
121
5
121
x+
= +
.
6
144
3
144
Step 3. Rewriting the left side and combining terms on the right side gives
x+
11
12
2
361
.
144
=
Step 4. Taking square roots, we get
11
x+
12
=
r
so
11
x=−
+
12
(b) We have a = 6, b = 11, c = −10, so
x=
which agrees with part (a).
−11 ±
361
144
√
361
12
and
and
11
x+
12
=−
11
x=−
−
12
r
361
,
144
√
361
.
12
√
√
(11)2 − 4 · 6(−10)
−11 ± 361
−11
361
=
=
±
,
2·6
12
12
12
p
9.3 SOLUTIONS
319
40. (a) We first move all the terms to the left side, 2x2 + 7x + 3 = 0. Then we divide both sides by 2, giving x2 + (7/2)x +
3/2 = 0.
Step 1. Add −3/2 to both sides, giving x2 + (7/2)x = −3/2.
Step 2. Since (7/2)/2 = 7/4 and (7/4)2 = 49/16, we add 49/16 to both sides:
x2 +
49
3
49
7
x+
=− +
.
2
16
2
16
Step 3. Rewriting the left side and combining terms on the right side gives
2
7
4
25
16
and
x+
25
.
16
=
Step 4. Taking square roots, we get
x+
7
4
=
r
x+
7
4
=−
r
25
,
16
so
7
5
1
7
5
+ =−
and x = − − = −3.
4
4
2
4
4
(b) We first move all the terms to the left side, 2x2 + 7x + 3 = 0. We have a = 2, b = 7, c = 3, so
x=−
x=
−7 ±
√
(7)2 − 4 · 2(3)
−7 ± 25
−7 ± 5
=
=
.
2·2
4
4
p
This gives x = −1/2 and x = −3, which agrees with part (a).
41. (a) We first move all the terms to the left side, 7x2 − x − 8 = 0. Then we divide both sides by 7, giving x2 + (−1/7)x −
8/7 = 0.
Step 1. Add 8/7 to both sides, giving x2 + (−1/7)x = 8/7.
Step 2. Since (−1/7)/2 = −1/14 and (−1/14)2 = 1/196, we add 1/196 to both sides:
x2 −
1
8
1
1
x+
= +
.
7
196
7
196
Step 3. Rewriting the left side and combining terms on the right side gives
x−
1
14
2
225
.
196
=
Step 4. Taking square roots, we get
1
x−
14
=
r
225
196
and
1
x−
14
=−
r
225
,
196
so
15
8
1
15
1
+
=
and x =
−
= −1.
14
14
7
14
14
(b) We first move all the terms to the left side, 7x2 − x − 8 = 0. We have a = 7, b = −1, c = −8, so
x=
x=
−(−1) ±
√
(−1)2 − 4 · 7(−8)
1 ± 225
1 ± 15
=
=
.
2·7
14
14
p
This gives x = 8/7 and x = −1, which agrees with part (a).
320
Chapter Nine /SOLUTIONS
42. (a) We first divide both sides by 4, giving x2 + x + 1/4 = 0.
Step 1. Add −1/4 to both sides, giving x2 + x = −1/4.
Step 2. Since (1)/2 = 1/2 and (1/2)2 = 1/4, we add 1/4 to both sides:
1
1
1
=− + .
4
4
4
Step 3. Rewriting the left side and combining terms on the right side gives
x2 + x +
2
1
2
=0
and
x+
= 0.
Step 4. Taking square roots, we get
x+
so
1
2
x+
x=−
1
2
= 0,
1
2
is the only solution.
(b) We have a = 4, b = 4, c = 1, so
p
(4)2 − 4 · 4(1)
−4 ±
=
2·4
8
This gives x = −1/2 as the only solution, which agrees with part (a).
x=
−4 ±
√
0
.
43. (a) We first divide both sides by 9, giving x2 − (6/9)x + 1/9 = 0, or x2 − (2/3)x + 1/9 = 0.
Step 1. Add −1/9 to both sides, giving x2 − (2/3)x = −1/9.
Step 2. Since (−2/3)/2 = −1/3 and (−1/3)2 = 1/9, we add 1/9 to both sides:
2
1
1
1
x+ =− + .
3
9
9
9
Step 3. Rewriting the left side and combining terms on the right side gives
x2 −
2
1
3
=0
and
x−
= 0.
Step 4. Taking square roots, we get
x−
so
1
3
x=
is the only solution.
(b) We have a = 9, b = −6, c = 1, so
x−
1
3
= 0,
1
3
√
(−6)2 − 4 · 9(1)
6± 0
=
.
2·9
18
This gives x = 1/3 as the only solution, which agrees with part (a).
x=
−(−6) ±
p
44. (a) We first divide both sides by 4, giving x2 + x + 3/4 = 0.
Step 1. Add −3/4 to both sides, giving x2 + x = −3/4.
Step 2. Since (1)/2 = 1/2 and (1/2)2 = 1/4, we add 1/4 to both sides:
3
1
1
=− + .
4
4
4
Step 3. Rewriting the left side and combining terms on the right side gives
x2 + x +
x+
1
2
2
1
=− .
2
Step 4. We cannot take the square root of −1/2 so this quadratic equation has no solutions.
9.3 SOLUTIONS
321
(b) We have a = 4, b = 4, c = 3, so
x=
We cannot calculate
√
−4 ±
√
(4)2 − 4 · 4(3)
−4 ± −32
=
.
2·4
8
p
−32, so this quadratic equation has no solutions, which agrees with part (a).
45. (a) We first divide both sides by 9, giving x2 − (6/9)x + 2/9 = 0, or x2 − (2/3)x + 2/9 = 0.
Step 1. Add −2/9 to both sides, giving x2 − (2/3)x = −2/9.
Step 2. Since (−2/3)/2 = −1/3 and (−1/3)2 = 1/9, we add 1/9 to both sides:
x2 −
1
2
1
2
x+ =− + .
3
9
9
9
Step 3. Rewriting the left side and combining terms on the right side gives
x−
1
3
2
1
=− .
9
Step 4. We cannot take the square root of −1/9 so this quadratic equation has no solutions.
(b) We have a = 9, b = −6, c = 2, so
p
√
−(−6) ± (−6)2 − 4 · 9(2)
6 ± −36
x=
=
.
2·9
18
√
We cannot calculate −36, so this quadratic equation has no solutions, which agrees with part (a).
46. We have a = 3, b = −2, c = −4, so:
x=
−(−2) ±
2±
√
p
(−2)2 − 4 · 3(−4)
2·3
52
.
6
√
√
Using a calculator, the solutions are x = (2 − 52)/6 = −0.8685 and x = (2 + 52)/6 = 1.5352. Note that by
factoring, we can write
√
√
√
52 = 4 · 13 = 2 13,
=
so another way to write the solutions is
√
√
1 ± 13
2 ± 2 13
=
.
x=
6
3
47. We have a = 5, b = −2, c = 2. The discriminant is
b2 − 4ac = (−2)2 − 4 · 5 · 2 = 4 − 40 = −36.
Since the discriminant is negative, there are no solutions.
48. We have a = 2, b = 7, c = 3, so the discriminant is
b2 − 4ac = (7)2 − 4 · 2(3) = 25.
This is positive, so there are two solutions.
49. We have a = 7, b = −1, c = −8, so the discriminant is
b2 − 4ac = (−1)2 − 4 · 7(−8) = 225.
This is positive, so there are two solutions.
50. We have a = 9, b = −6, c = 1, so the discriminant is
b2 − 4ac = (−6)2 − 4 · 9(1) = 0,
so there is one solution.
322
Chapter Nine /SOLUTIONS
51. We have a = 4, b = 4, c = 1, so the discriminant is
b2 − 4ac = (4)2 − 4 · 4(1) = 0,
so there is one solution.
52. We have a = 9, b = −6, c = 2, so the discriminant is
b2 − 4ac = (−6)2 − 4 · 9(2) = −36.
This is negative, so there are no solutions.
53. We have a = 4, b = 4, c = 3, so the discriminant is
b2 − 4ac = (4)2 − 4 · 4(3) = −32.
This is negative, so there are no solutions.
PROBLEMS
54. (a) Since at t = 0 we have
Height = h(0) = −16 · 02 + 900 = 900,
the sunglasses were dropped from a height of about 900 feet. (The third and highest platform of the Eiffel Tower is
in fact 906 feet above the ground.)
(b) The glasses hit the ground when the height is zero, so we must solve the equation −16t2 + 900 = 0. Adding 16t2 to
both sides we get
900 = 16t2
900
= t2
16
r
t=±
divide both sides by 16
take the square root of each side.
900
30
15
=±
=±
= ±7.5.
16
4
2
Since the formula for the height only applies for t ≥ 0 (after the sunglasses are dropped), the sunglasses hit the
ground at about t = 7.5 seconds, or 7.5 seconds after they are dropped.
55. After the ball has fallen k feet, its height above the ground is 100 − k, so we have to solve the equation
100 − k = −16t2 + 100
−k = −16t2
t=
r
√
k
k
=±
.
16
4
Since the time taken to fall must be a positive number, we choose the positive square root, t =
√
k/4.
56. Since the left-hand side is always positive or zero, this equation has no solution when A is negative.
57. This equation has two solutions for all values of A, so there are no conditions on A for which the equation has no solution.
58. When we subtract 5 from both sides, we have A(x − 2)2 = −5. Since (x − 2)2 is always positive or zero, this equation
has no solution when A is positive since we would have a positive expression equal to a negative solution. It also has no
solution when A = 0. This equation has no solution when A ≥ 0.
59. If we subtract A from both sides, we have 5(x − 3)2 = 10 − A. Since the left-hand side 5(x − 3)2 is always positive or
zero, this equation has no solution when 10 − A is negative, which is when A > 10. Alternately, we can reason that the
expression on the left is always greater than or equal to A so it will never equal 10 if A > 10. With either reasoning, we
conclude that the equation has no solution when A > 10.
9.3 SOLUTIONS
323
60. No. The second equation is quadratic, and we can solve it by factoring:
x2 − 2x − 3 = 0
(x − 3)(x + 1) = 0,
so the solutions are x = 3 and x = −1. However, x = −1 is not a solution to the original first equation:
p
√
2(−1) + 3 = 1 6= −1.
61. (a) The area of the window is obtained by adding the area of the rectangle to the area of the semicircle,
A = lw +
πr 2
.
2
Since the diameter of the circle is equal to the width of the rectangle, the radius is half the width of the rectangle.
2
= lw + πw2 /8.
Thus, A = lw + π(w/2)
2
(b) To find the dimensions of the rectangle, we let l = 2w, set the equation equal to 20 and solve.
πw2
8
πw2
2w(w) +
8
16w2 + πw2
8
16 + π
w2
8
8.358
A = lw +
20 =
20 =
20 =
w2 =
w = 2.891.
To the nearest tenth, the width is 2.9 feet, the length is 5.8 feet, and the radius of the circle is 1.5 feet.
62. (a) Since the arch is at its highest in the center of the bridge, to find the height of the arch, we set x = 0. When
x = 0, h(x) = 876. Thus, the arch is 876 feet high.
(b) To find the span of the arch, we set the height equal to 575 and solve.
h(x) = −0.00121246x2 + 876
575 = −0.00121246x2 + 876
0.00121246x2 = 301
301
0.00121246
x2 = 248,255.613
x2 =
x = 498.253.
If, at 575 feet above the ground, the arch of the bridge extends 498.253 feet to each side of the center, its entire span
at that point would be approximately 997 feet.
63. If we let w be the width of the rectangle, the length is w + 2. When we cut one inch from each end of the rectangle to
form the box, the width of the box becomes (w − 2) and the length becomes (w + 2 − 2) = w. The height of the box is
1. We can substitute this information into the formula for volume and solve:
V = lwh
80 = (w)(w − 2)(1)
80 = w2 − 2w
80 + 1 = w2 − 2w + 1
324
Chapter Nine /SOLUTIONS
81 = (w − 1)2
√
w − 1 = ± 81
w = 1±9
w = 10
or
w = −8.
Since we are dealing with dimensions of a rectangle, we are only interested in the positive answer. Thus, the dimensions
of the paper are 10 inches by 12 inches.
64. (a) We have
302
= 111 ft.
20
602
At 60 mph, stopping distance = d = 2.2(60) +
= 312 ft.
20
2
90
At 90 mph, stopping distance = d = 2.2(90) +
= 603 ft.
20
At 30 mph, stopping distance = d = 2.2(30) +
(b) From our answers in part (a), we see that the speed for a stopping distance of 500 ft is between 60 mph and 90 mph.
Figure 9.3 shows a graph of d = 2.2v + v 2 /20. We see that the speed corresponding to a stopping distance of 500
ft is approximately 80 mph. To find the speed more accurately, we use the quadratic formula to solve the quadratic
equation:
1 2
v2
which is
v + 2.2v − 500 = 0,
500 = 2.2v +
20
20
giving v = 80.39 and v = −124.39. On physical grounds we want v > 0, so v = 80.39, which is in good agreement
with our graphical estimate of 80 mph.
d (ft)
(80, 500)
500
30
60
v (mph)
90
Figure 9.3
65. The key thing to realize here is that if a and c have opposite signs, then one value must be positive and the other negative.
(a) There are two possibilities for the graph:
(i) a is positive and c is negative: In this case, the graph of y = ax2 + bx + c is an upward-opening parabola with
a negative y-intercept. Thus, the graph must have two x-intercepts.
(ii) a is negative and c is positive: In this case, the graph of y = ax2 + bx + c is a downward-opening parabola with
a positive y-intercept. Once again, the graph must have two x-intercepts.
Since in either case the graph has two x-intercepts, the equation ax2 + bx + c = 0 has two solutions.
(b) We know that ac is negative because the product of a positive and a negative is negative. We know that b2 is never
negative, so
b2 − 4ac = Non-negative −4 · (Negative) = Positive.
|
{z
b2
}
|
Since the discriminant is positive, the equation has two solutions.
{z
ac
}
9.3 SOLUTIONS
66. The equations are graphed in Figure 9.4. To find the x-intercepts, we solve f (x) = 0.
(a) We solve x2 − 4x + 5 = 0 to get
−(−4) ±
x=
√
(−4)2 − 4(1)(5)
4 ± −4
=
.
2(1)
2
p
Since the discriminant is negative there are no real solutions. Thus the graph has no x-intercepts.
(b) We solve x2 − 4x + 4 = 0 to get
x=
−(−4) ±
√
(−4)2 − 4(1)(4)
4± 0
=
.
2(1)
2
p
Since the discriminant is zero, x = 2 is the only solution, and the graph has one x-intercept.
(c) We solve x2 − 4x + 3 = 0 to get
x=
−(−4) ±
√
p
(−4)2 − 4(1)(3)
2(1)
16 − 12
2
√
4± 4
=
= 3 and 1.
2
=
4±
Since the discriminant is positive there are two x-intercepts, 3 and 1.
(d) We solve x2 − 4x + 2 = 0 to get
x=
−(−4) ±
√
p
(−4)2 − 4(1)(2)
(2)(1)
16 − 8
2
√
4± 8
=
2
√
= 2 ± 2.
=
4±
Since the discriminant is positive, there are two x-intercepts, 2 +
y
√
2 and 2 −
8
(a) (b)
(c)
7
(d)
6
1
x
1
4
Figure 9.4
5
√
2.
325
326
Chapter Nine /SOLUTIONS
67. Here, a = 3, b = 2, and c is unknown. We know that in order for this equation to have two solutions, the discriminant
b2 − 4ac must be positive. This means
22 − 4 · 3c > 0
4 − 12c > 0
12c < 4
4
c<
12
1
c< .
3
This means c must be less than 1/3 in order for the equation to have two different solutions.
68. Here, a = 2, c = 8, and b is unknown. We know that in order for this equation to have two solutions, the discriminant
b2 − 4ac must be positive. This means
b2 − 4 · 2 · 8 > 0
b2 − 64 > 0
b2 > 64.
For b2 to be greater than 64, b must be greater than 8 or less than −8. Thus, we conclude that for there to be two solutions,
either b > 8 or b < −8.
69. If a 6= 0, we have a quadratic. The discriminant of 2ax2 − 2(a − 1)x − 1 = 0 is
(−2(a − 1))2 − 4 · (2a) · (−1) = 4(a − 1)2 + 8a = 4a2 + 4,
which is always positive. This means that the quadratic equation has two solutions.
If a = 0 the equation 2ax2 − 2(a − 1)x − 1 = 0 becomes 2x − 1 = 0, so x = 1/2 is the only solution. The equation
is not quadratic in this case.
70. To find the common x values, we substitute y = −x + b into y = c/x, and find −x + b = c/x, which is the quadratic
−x2 + bx − c = 0 with discriminant b2 − 4(−1)(−c) = b2 − 4c.
(a) The curves will have no points in common if the quadratic −x2 + bx − c = 0 has no real solutions, which occurs
when the discriminant is negative, that is b2 − 4c < 0.
(b) The curves will have exactly one point in common if the quadratic −x2 + bx − c = 0 has exactly one solution, which
occurs when the discriminant is zero, that is b2 − 4c = 0.
(c) The curves will have exactly two points in common if the quadratic −x2 + bx − c = 0 has two solutions, which
occurs when the discriminant is positive, that is b2 − 4c > 0.
(d) A quadratic equation has at most two solutions, so it is not possible for the curves to have more than two points in
common.
71. If the equation has a real solution, then the discriminant b2 − 4ac must not be negative:
2
b2 − 4ac ≥ 0
b − 4 · 2 · 50 ≥ 0
b2 − 400 ≥ 0
b2 ≥ 400.
We see that the square of b must be larger than 400, so b must be no less than 20 or no greater than −20, that is b ≤ −20
or b ≥ 20.
72. One approach is to apply the quadratic formula. We know that the largest solution is given by
√
−3 + 32 − 4 · 1 · c
.
x=
2·1
9.4 SOLUTIONS
Since the largest solution is x = 3, we have
−3 +
√
32 − 4 · 1 · c
=3
2 √
−3 + 9 − 4c = 6
√
9 − 4c = 9
9 − 4c = 81
−4c = 72
c = −18.
To check our answer, we see that
x2 + 3x − 18 = (x − 3)(x + 6).
This means that x = −6, 3 are the solutions to the equation x2 + 3x − 18 = 0, and that x = 3 is the largest.
73. According to the quadratic formula, one solution to the equation ax2 + bx + c = 0 is given by
√
−b + b2 − 4ac
x=
.
2a
Since
√
−2 ± 8
,
2
we can try letting −b = −2, so b = 2, and a = 1. This means the discriminant, b2 − 4ac, must equal 8:
x=
b2 − 4ac = 8
22 − 4 · 1 · c = 8
4 − 4c = 8
−4c = 4
c = −1.
Thus, one possible equation is
x2 + 2x − 1 = 0.
Solutions for Section 9.4
EXERCISES
1. From the zero-factor principle, we see that x = 2, 3.
2. From the zero-factor principle, we see that x = 0, 5.
3.
x2 + 5x + 6 = 0
(x + 2)(x + 3) = 0
x = −2, −3.
327
328
Chapter Nine /SOLUTIONS
4. Noting that the sum of 9 and 4 is 13 and the product is 36, we see that
6x2 + 13x + 6 = 0
6x2 + 9x + 4x + 6 = 0
3x(2x + 3) + 2(2x + 3) = 0
(3x + 2)(2x + 3) = 0,
5.
so 3x + 2 = 0 which gives x = −2/3, and 2x + 3 = 0 which gives x = −3/2.
(x − 1)(x − 3) = 8
x2 − 3x − x + 3 = 8
x2 − 4x − 5 = 0
(x − 5)(x + 1) = 0,
6.
so x = 5, −1.
(2x − 5)(x − 2)2 = 0,
so 2x − 5 = 0 which gives x = 5/2, or x − 2 = 0 which gives x = 2. Thus, x = 5/2, 2.
7. We can solve this by rewriting it in the form of a quadratic:
x4 + 2x2 + 1 = 0
(x2 )2 + 2x2 + 1 = 0
(x2 + 1)(x2 + 1) = 0.
By the zero-factor principle, this equation is true provided
x2 + 1 = 0
x2 = −1.
This equation has no solutions, so there are no solutions to the original equation.
8.
x4 − 1 = 0
(x2 − 1)(x2 + 1) = 0
x = ±1.
9. By the zero-factor principle, x = 3, −2, −7.
10.
x(x2 − 4)(x2 + 1) = 0
x(x − 2)(x + 2)(x2 + 1) = 0,
so x = 0, 2, −2. Note that the expression x2 + 1 can’t be further factored, and that its value is positive for all x.
11. The quadratic expression on the left-hand side is set equal to zero and in factored form, so we set each factor equal to zero
and see that the two solutions are x = 3 and x = −5.
12. We solve this either by rewriting the equation as x2 = 4 and then taking square roots, or by factoring as (x − 2)(x + 2).
Either way, we see that the two solutions are x = −2 and x = 2.
9.4 SOLUTIONS
329
13. The quadratic expression on the left-hand side is in factored form and set equal to zero, so we set each factor equal to zero
and see that the solutions are x = 0 and x = −2.
14. To solve a quadratic expression such as this, we first multiply out and set it equal to zero:
x(x + 3) = 10
x2 + 3x = 10
2
x + 3x − 10 = 0.
If possible, factor the quadratic expression. In this case, the expression factors as follows:
(x + 5)(x − 2) = 0.
We set each factor equal to zero to see that the solutions are x = −5 and x = 2.
15. To solve a quadratic expression such as this, we first move all the terms to the left and then combine like terms:
x2 + 2x = 5x + 4
x2 + 2x − 5x − 4 = 0
x2 − 3x − 4 = 0.
If possible, factor the quadratic expression. In this case, the expression factors as follows:
(x − 4)(x + 1) = 0.
We set each factor equal to zero to see that the solutions are x = 4 and x = −1.
16. The quadratic expression is set equal to zero, so we factor it if possible. We then set each factor equal to zero.
2x2 + 5x = 0
x(2x + 5) = 0
x=0
2x + 5 = 0
x = −5/2.
The two solutions are x = 0 and x = −5/2.
17. The quadratic expression is set equal to zero, so we factor it if possible. We then set each factor equal to zero.
x2 − 8x + 12 = 0
(x − 2)(x − 6) = 0.
The two solutions are x = 2 and x = 6.
18. The quadratic expression is set equal to zero, so we factor it if possible. In this case, the expression does not appear to
factor easily so we use the quadratic formula with a = 1, b = 3 and c = 7:
√
−b ± b2 − 4ac
x=
2a
p
−3 ± 32 − 4(1)(7)
=
2
√
−3 ± 9 − 28
=
.
2
Since the discriminant is negative, we see that this equation has no solutions.
330
Chapter Nine /SOLUTIONS
19. The quadratic expression is set equal to zero, so we factor it if possible. In this case, the expression does not appear to
factor easily so we use the quadratic formula with a = 1, b = 6, and c = −4:
√
−b ± b2 − 4ac
x=
2a
p
−6 ± 62 − 4(1)(−4)
=
2
√
−6 ± 36 + 16
=
2
√
−6 ± 52
=
.
2
√
√
There are two solutions: x = (−6 − 52)/2 = −6.606 and x = (−6 + 52)/2 = 0.606.
20. The quadratic expression is set equal to zero, so we factor it if possible. In this case, the expression does factor and we
then set each factor equal to zero and solve:
2x2 − 5x − 12 = 0
(2x + 3)(x − 4) = 0
2x + 3 = 0
x = −3/2
x−4 =0
x = 4.
Alternatively, we can use the quadratic formula with a = 2, b = −5 and c = −12:
√
−b ± b2 − 4ac
x=
2a
p
−(−5) ± (−5)2 − 4(2)(−12)
=
2(2)
√
5 ± 25 + 96
=
4
√
5 ± 121
=
4
5 ± 11
=
.
4
There are two solutions: x = (5 − 11)/4 = −6/4 = −1.5 and x = (5 + 11)/4 = 16/4 = 4.
Of course, whether we factor or use the quadratic formula, we get the same answer. The solutions are x = −1.5 and
x = 4.
21. We start by moving all the terms to the left:
x2 − 3x + 12 = 5x + 5
x2 − 3x + 12 − 5x − 5 = 0
x2 − 8x + 7 = 0.
We can factor the quadratic expression or we can use the quadratic formula. In this case, the expression factors, and we
set each factor equal to zero:
(x − 1)(x − 7) = 0.
The two solutions are x = 1 and x = 7.
22. We begin by multiplying out and moving all the terms to the left:
2x(x + 1) = 5(x − 4)
2x2 + 2x = 5x − 20
2x2 + 2x − 5x + 20 = 0
2x2 − 3x + 20 = 0.
9.4 SOLUTIONS
331
The quadratic expression on the left-hand side does not appear to factor easily, so we use the quadratic formula with
a = 2, b = −3 and c = 20:
x=
=
−(−3) ±
3±
√
p
(−3)2 − 4(2)(20)
2(2)
9 − 160
.
4
Since the discriminant is negative, there are no solutions.
23. Given a = 3, b = −2, c = −11, we first find the discriminant:
b2 − 4ac = (−2)2 − 4 · 3(−11) = 4 − (−132) = 136.
This is positive, so there are two solutions:
√
b2 − 4ac
2a
√
−(−2) ± 136
=
2·3
√
2 ± 136
=
.
6
x=
This can be simplified by writing
√
−b ±
136 =
√
giving
√
4 · 34 = 2 34,
√
2 ± 2 34
x=
6
√
1 ± 34
.
=
3
Approximate numerical values for x are x = −1.610, 2.277.
24. Given a = 5, b = 3, c = 3, we first find the discriminant:
b2 − 4ac = 32 − 4 · 5 · 3
= 9 − 60
= −51.
This is negative, so there are no solutions.
25. This is in factored form. We have the solutions
2x − 3 = 0
3
x= ,
2
and
3x − 1 = 0
1
x= .
3
332
Chapter Nine /SOLUTIONS
26. This function is in standard form with a = 3, b = −5, c = −1. The discriminant is
b2 − 4ac = (−5)2 − 4(3)(−1) = 25 + 12 = 37.
Since the discriminant is positive, there are two real zeros:
√
−(−5) ± 37
x=
2·3
√
5 ± 37
=
.
6
27. We factor:
y = x(6x − 10) − 7(3x − 5)
= 2x(3x − 5) − 7(3x − 5)
= (2x − 7)(3x − 5).
Setting this equal to zero and using the zero-factor principle, we have
And
2x − 7 = 0
7
x=
2
3x − 5 = 0
5
x= ,
3
so the zeros are x = 7/2, 5/3.
√
√
28. We have (x − 5)(x + 5) = 0 so x2 = 5.
29. (x − 2)(x + 3) = 0.
30. x(2x − 3) = 0.
31. (x − 2)2 = 3.
32. x(x + p) = 0.
33. We have
√
√
q)) · (x − (p − q)) = 0
√
√
((x − p) − q) · ((x − p) + q) = 0
(x − (p +
(x − p)2 − q = 0
(x − p)2 = q.
34. We have (x − a)(x − b) = 0.
35. We have
2
(x − a)2 = 0
x − 2ax + a2 = 0.
36. x2 + 1 = 0 has no solutions, because x2 is always non-negative, so x2 + 1 is always greater than or equal to 1. Other
answers are possible.
9.4 SOLUTIONS
333
37. Factoring out an x we get
x(x − x2 + 2) = 0
x(−x2 + x + 2) = 0
−x(x2 − x − 2) = 0
−x(x − 2)(x + 1) = 0,
so the solutions are x = 0, x = 2, and x = −1.
38. We can view this as a quadratic equation in x2 :
−2(x2 ) − 3 + (x2 )2 = 0.
Rearranging and factoring we get
(x2 )2 − 2(x2 ) − 3 = 0
(x2 − 3)(x2 + 1) = 0.
√
√
2
Now x2 − 3 is zero when
√ x = 3 or − 3 and x + 1 is never zero because it is always greater than or equal to 1, so the
solutions are x = ± 3.
1
39. If we multiply x + = 2 by x, the equation becomes x2 + 1 = 2x, or x2 − 2x + 1 = 0. This is a quadratic equation in
x
x which can be factored, (x − 1)(x − 1) = 0, with solution x = 1.
40. Multiplying both sides by y(y − 3) means multiplying all three terms on the left:
2
2
· y(y − 3) −
· y(y − 3) − 3y(y − 3) = 0.
y
y−3
Canceling gives
2(y − 3) − 2y − 3y(y − 3) = 0.
To solve, we first expand and collect terms, giving
2y − 6 − 2y − 3y 2 + 9y = 0
−3y 2 + 9y − 6 = 0.
Factoring this quadratic, we see
−3(y 2 − 3y + 2) = 0
−3(y − 1)(y − 2) = 0,
so the solutions are y = 1 and y = 2. Since these solutions do not make the denominator of the original equation zero,
these are the solutions to the original equation.
41. Multiplying both sides by z 2 − 4 gives
12
3
· (z 2 − 4) − 2
· (z 2 − 4) = 1 · (z 2 − 4).
z−2
z −4
Since z 2 − 4 = (z − 2)(z + 2), canceling produces
3(z + 2) − 12 = z 2 − 4.
To solve, we set the equation equal to zero, expand, and collect terms.
3(z + 2) − 12 − z 2 + 4 = 0
3z + 6 − 12 − z 2 + 4 = 0
−z 2 + 3z − 2 = 0
334
Chapter Nine /SOLUTIONS
Factoring this quadratic, we get
−1(z 2 − 3z + 2) = 0
−1(z − 1)(z − 2) = 0.
This means
z=1
and
z = 2.
However, z = 1 is the only solution to the original equation. Since the denominator of the original equation is zero when
z = 2, it is an extraneous solution.
42. Letting x = t2 , we have x2 = t4 , and we can write
x2 − 13x + 36 = 0
(x − 4)(x − 9) = 0,
so x = 4 and x = 9. Substituting x = t2 , we have
t2 = 4
t = ±2
t2 = 9
and
t = ±3,
so t = −3, −2, 2, 3.
43. Letting x = t2 , we have x2 = t4 , and we can write
x2 − 3x − 10 = 0
(x − 5)(x + 2) = 0,
so x = 5 and x = −2. Substituting x = t2 , we have
t2 = 5
√
t = ± 5.
√
The other solution, x = −2, does not correspond to solutions to t, since t2 = −2 has no solutions. Thus, t = ± 5.
√
44. Letting x = t, we have x2 = t, and can write
x2 + 2x − 15 = 0
Substituting x =
√
(x + 5)(x − 3) = 0.
t gives
√
√
t+5
t − 3 = 0,
√
√
√
so t = −5 and t = 3. The first equation has no solutions, since t is either positive or undefined for all values of t.
The second equation has t = 9 for a solution, so the only solution to the original equation is t = 9.
45. Letting x = (t − 3)3 , we have x2 = (t − 3)3
2
= (t − 3)6 , and we can write
x2 − 5x + 6 = 0
(x − 2)(x − 3) = 0,
3
so x = 2 or x = 3. Substituting x = (t − 3) gives
(t − 3)3 = 2
√
3
t−3 = 2
9.4 SOLUTIONS
t = 3+
and
(t − 3)3 = 3
√
3
t−3 = 3
t = 3+
so t = 3 +
√
3
2 or t = 3 +
√
3
√
3
2
√
3
3,
335
3.
46. (a) We set each factor equal to zero to see that the solutions are x = 3 and x = −2.
(b) To use the quadratic formula, we first multiply out to write the equation in the form ax2 + bx + c = 0.
(x − 3)(x + 2) = 0
x2 − 3x + 2x − 6 = 0
x2 − x − 6 = 0.
We use the quadratic formula with a = 1, b = −1 and c = −6:
p
(−1)2 − 4(1)(−6)
2
p
1 ± 1 − (−24)
=
2
√
1 ± 25
=
2
1±5
.
=
2
x=
−(−1) ±
There are two solutions: x = (1 − 5)/2 = −4/2 = −2 and x = (1 + 5)/2 = 6/2 = 3. As expected, these are the
same solutions we found by factoring.
47. (a) We factor the equation and set each factor equal to zero:
x2 + 7x + 12 = 0
(x + 3)(x + 4) = 0.
The solutions are x = −3 and x = −4.
(b) We use the quadratic formula with a = 1, b = 7 and c = 12:
√
−b ± b2 − 4ac
x=
2a
p
−7 ± 72 − 4(1)(12)
=
2
√
−7 ± 49 − 48
=
2
√
−7 ± 1
=
2
−7 ± 1
=
.
2
There are two solutions: x = (−7 − 1)/2 = −8/2 = −4 and x = (−7 + 1)/2 = −6/2 = −3. As expected, these
are the same solutions we found by factoring.
PROBLEMS
48. Multiplying by denominators in the first four equations gives
Equation I:
(2x − 1)(x + 3) = x − 3
336
Chapter Nine /SOLUTIONS
(2x − 1)(x − 3) = x + 3
Equation II:
(x − 3)(x + 3) = 2x − 1
Equation III:
(x + 3)(x − 3) = 2x − 1.
Equation IV:
Thus equations I, V have the same solutions. Equations II, VII have the same solution. Equations III, IV, VI have the same
solution. None of these groups have the same solutions.
Before multiplying by the denominators we should check that this does not introduce any extraneous solutions. For
example, for Equation I, because we multiplied through by (x − 3)(x + 3), we should check that 3 and −3 are not
solutions to the new equation.
49. Two solutions. The expression on the left is in factored form, so we see that there are two solutions, at x = 3 and at
x = −2.
50. One solution. The expression on the left is in factored form, with the two factors the same. There is one solution, at x = 2.
51. No solutions. The expression on the left is in factored form, but it is not set equal to zero. The expression on the left is a
square so it can never be equal to a negative number.
√
52. Two solutions. We would begin to solve this by taking the square root of both sides. The resulting ± 17 on the right
would lead to two different solutions.
53. One solution. We can look at this in two ways. The expression (x − 3)(x − 3) on the left is in factored form with the
same factor appearing twice and is set equal to zero, so the
√ only solution is x = 3. Alternatively, we can begin to solve
this equation by taking the square root of both sides, and 0 = 0, so there is exactly one solution.
54. No solution. We can look at this in two ways. We know that the expression on the left has a smallest value of 5 since the
term 3(x + 2)2 is never negative. The expression on the left is never less than 5 so it cannot equal 1. Alternatively, we
can think through the initial step of solving this equation: when we subtract 5 from both sides, we have 3(x + 2)2 = −4.
Since the left side is always positive and the right side is negative, there is no solution.
55. Two solutions. We can look at this in two ways. We know that the expression on the left has a largest value of 7 since the
term −2(x − 1)2 is never positive. The term takes on two solutions for all negative numbers, so the entire expression on
the left is equal to all values less than 7 at two different values of x. Alternatively, we can think through the initial steps of
solving this equation: we subtract 7 from√both sides and then divide by −2, which gives us (x − 1)2 = 1. When we take
the square root of both sides, we have ± 1 on the right, which leads to two different solutions.
56. One solution. We can look at this in two ways. The expression on the left has a smallest value of 10 since the term
2(x − 3)2 is never negative. The expression is equal to 10 at exactly x = 3 which is the only solution. Alternatively, we
can think through the initial step of solving this equation: subtracting 10 from both sides gives us 2(x − 3)2 = 0, which
has one solution at x = 3.
57. (a) When the diver hits the water, s = 0, so we solve s(t) = −16t2 + 20t + 6 = 0
−16t2 + 20t + 6 = 0
−2(8t2 − 10t − 3) = 0 factoring out −2
−2(4t + 1)(2t − 3) = 0 factoring
Therefore,
4t + 1 = 0
or
2t − 3 = 0
t = −1/4
or
t = 3/2.
so
Since t ≥ 0, the only solution is t = 3/2. Therefore, the diver is in the air 1.5 sec before he hits the water.
(b) The height function is quadratic and the coefficient of the t2 term is negative, so its graph is a parabola that opens
downward. Finding the coordinates of its vertex, (t, s) will produce the maximum height s, and the time, t, at which
this maximum is reached. Since the constant term is not zero, we first subtract 6 from both sides and then complete
the square.
s = −16t2 + 20t + 6
s − 6 = −16t2 + 20t
subtracting 6 from both sides
9.4 SOLUTIONS
s−6
−16
s−6
−16
s−6
−16
s−6
−16
25
+
64
25
+
64
s−6
−16
20
t
16
5
2
t − t
4
5
25
t2 − t +
4 2 64
5
t−
8
5 2 25
−
t−
8 5 64
2
25
−16 t −
+
+6
8
4
= t2 −
dividing by −16
=
reducing
=
=
=
s=
s = −16 t −
5
8
2
49
4
+
337
completing the square
rewriting the right side
subtracting
25
64
from both sides
multiply by −16 and adding 6
to both sides
combining like terms
This means that the vertex (t, s) is ( 85 , 49
), so the maximum height is
4
reached this height is 58 = 0.625 sec after the jump began.
49
4
= 12.25 ft, and the time the diver
58. (a) The ball is on the ground when t = 0, so its height is 0 when t = 0. Putting t = 0 in the expression for height, we
get
Height = −4.9(0)2 + 30(0) + c = c meters,
so c = 0.
(b) The factored form of the expression for height makes it easy to find the zeros. Factoring out −4.9t gives
30
= −4.9t(t − 6.122).
4.9
The only values of t that make the height equal to zero are t = 0 seconds, which corresponds to the time the ball
began its flight, and t = 6.122 seconds, which is the time the ball hits the ground again.
−4.9t2 + 30t = −4.9t t −
59. The original area is 4 · 6 = 24 square feet. The new area is 48 square feet. After adding a strip of uniform width x to each
of the four sides, the new dimensions are 2x + 4 and 2x + 6. We enter this information into the area formula and solve:
A = lw
48 = (2x + 4)(2x + 6)
48 = 4x2 + 20x + 24
0 = 4x2 + 20x − 24
0 = x2 + 5x − 6
0 = (x + 6)(x − 1)
x = −6 or
x = 1.
Since we are dealing with dimensions of a rectangle, we reject the negative value. The strip should be 1 foot wide
60. If we multiply x−1 +2−1 = (x+2)−1 by 2x(x+2), the equation becomes 2(x+2)+x(x+2) = 2x, or x2 +2x+4 = 0.
This is a quadratic equation whose discriminant, b2 − 4ac = 22 − 4 · 1 · 4 = −12, is negative, so it has no solutions.
61. (a) This expression is quadratic in p, so we use the quadratic formula with a = 1, b = 2q and c = 5q:
p
(2q)2 − 4(1)(5q)
p 2
−2q ± 4q 2 − 20q
=
2
p
−2q ± 2 q 2 − 5q
=
2
p=
The solutions are p = −q +
p
−2q ±
= −q ±
p
q 2 − 5q and p = −q −
q 2 − 5q.
p
q 2 − 5q.
338
Chapter Nine /SOLUTIONS
(b) The equation is linear in q, so we put constant terms (without a q) on one side and terms with q on the other side:
p2 + 2pq + 5q = 0
2pq + 5q = −p2
(2p + 5)q = −p2
q=
−p2
.
2p + 5
There is one solution, as we expect with a linear equation.
62. (a) The equation is linear in p, so we put constant terms (without a p) on one side and terms with p on the other side:
q 2 + 3pq = 10
3pq = 10 − q 2
10 − q 2
p=
.
3q
There is one solution, as we expect with a linear equation.
(b) This expression is quadratic in q, so we first move all the terms to the left:
q 2 + 3pq − 10 = 0
and then use the quadratic formula with a = 1, b = 3p and c = −10:
p
(3p)2 − 4(1)(−10)
p 2
−3p ± 9p2 + 40
.
=
2
q=
The solutions are q = (−3p −
p
−3p ±
9p2 + 40)/2 and q = (−3p +
p
9p2 + 40)/2.
63. (a) This expression is quadratic in p, so we use the quadratic formula with a = q 2 , b = −1 and c = 2:
p=
=
p
−(−1) ±
1±
p
(−1)2 − 4(q 2 )(2)
2q 2
p
1 − 8q 2
.
2q 2
p
The solutions are p = (1 − 1 − 8q 2 )/(2q 2 ) and p = (1 + 1 − 8q 2 )/(2q 2 ).
(b) This expression is quadratic in q. We can either use the quadratic formula with a = p2 , b = 0 and c = −p + 2, or
we can solve for q 2 and take square roots. Either way, we see that
√
p−2
q=±
.
p
64. This expression is quadratic in both variables. Since the quadratic expression factors in an obvious way, we pull out the
common factors and set each factor equal to zero to solve for the appropriate variable:
pq 2 + 2p2 q = 0
pq(q + 2p) = 0
(a) To solve for p, we set each factor equal to zero and solve for p. We see that the solutions are p = 0 and p = −q/2.
(b) To solve for q, we set each factor equal to zero and solve for q. We see that the solutions are q = 0 and q = −2p.
9.5 SOLUTIONS
339
65. (a) In the discriminant formula, we have c = 0 so b2 − 4ac = b2 and we see that the discriminant is positive or zero.
Since the discriminant is not negative, the equation has solutions.
(b) We factor the equation and set each factor equal to zero:
ax2 + bx = 0
x(ax + b) = 0
x=0
ax + b = 0
x = −b/a.
The solutions are x = 0 and x = −b/a.
(c) We use the quadratic formula with c = 0:
√
b2 − 4ac
2a
√
−b ± b2 − 0
=
2a
√
−b ± b2
=
2a
−b ± b
=
.
2a
x=
−b ±
The solutions are x = (−b − b)/(2a) = −2b/(2a) = −b/a and x = (−b + b)/(2a) = 0/(2a) = 0. As expected,
these are the same solutions we found by factoring.
66. No. If a product is zero, at least one its factors must be zero. However, if a product is 6, one of its factors does not have to
be 6. For example, the factors could be 2 and 3.
67. The student is wrong. You do not solve a quadratic such as x(x + 1) = a · b by setting x = a and x + 1 = b. On the other
hand, solving x(x + 1) = 0 does lead to x = 0 and x + 1 = 0.
The quadratic x(x+1) = 2·6 can be rewritten as x2 +x−12 = 0, which can be factored in the form (x−3)(x+4) =
0. Thus, the solutions are x = 3 and x = −4.
68. Checking the solutions z = 16 and z = 4, we see that:
√
16 − 2 16 = 8, the solution z = 16 is valid
√
4 − 2 4 6= 8, the solution z = 4 is invalid.
√
The problem with the given derivation is that the original equation, z − 2 z = 8, is not algebraically equivalent to the
resulting equation, z 2 − 20z + 64 = 0. This is because when we squared both sides of the equation in the step (3), we
obtained an equation that has more solutions than the original. To see how this can happen in a simpler case, notice that
the equation 2x = 4 has only one solution, but when we square both sides, the resulting equation, 4x = 16, has two
solutions.
69. No. Although both these expressions equal zero at x = 3/2 and x = 7/5, they don’t have the same value elsewhere. For
instance, at x = 0 we have
2
2
29x + 21
10x −
= 10
·0 −
29
· 0 +21 = 21
x−
3
2
x−
7
5
= 0−
3
2
Solutions for Section 9.5
EXERCISES
1. (2 − 6i) + (23 − 14i) = (2 + 23) + (−6i − 14i) = 25 − 20i.
0−
7
5
=
21
.
10
340
Chapter Nine /SOLUTIONS
2. (7 + i)(2 − 5i) = 14 − 35i + 2i − 5i2 = 14 − 33i + 5 = 19 − 33i.
5(4 − 3i)
5(4 − 3i)
5(4 − 3i)
5
4 − 3i
4 − 3i
4
3
5
=
·
=
=
=
=
= − i
3.
4 + 3i
4 + 3i 4 − 3i
16 − 9i2
16 + 9
25
5
5
5
√
√
√ √
4. −12 = −4 3 = 2i 3
√
√
5. −81 − −4 = 9i − 2i = 7i
√
√
6. 3 −25 − 4 −64 = 3(5i) − 4(8i) = 15i − 32i = −17i
√
√
7. 2 −16 − 5 −1 = 2(4i) − 5i = 8i − 5i = 3i
8. (−4i3 )(−2i) + 6(5i3 ) = 8i4 + 30i3 = 8(1) + 30(−i) = 8 − 30i
9. (7 + 5i) − (12 − 11i) = (7 + 5i) + (−12 + 11i) = (7 − 12) + (5i + 11i) = −5 + 16i
10. 6(3 + 4i) − 2i(i + 5) = 18 + 24i − 2i2 − 10i = 18 + 24i + 2 − 10i = 20 + 14i
11. (5 + 2i)2 = 25 + 20i + 4i2 = 25 + 20i − 4 = 21 + 20i
1
1
4 + 5i
4 + 5i
4 + 5i
4 + 5i
4
5
12.
=
·
=
=
=
=
+
i
4 − 5i
4 − 5i 4 + 5i
16 − 25i2
16 + 25
41
41
41
13. (3 + 8i) + 2(4 − 7i) = 3 + 8i + 8 − 14i = 11 − 6i
14. (9 − 7i) − 3(5 + i) = 9 − 7i − 15 − 3i = −6 − 10i
√
√
√
√
√
√ √
√ √
15. 3 −5 + 5 −45 = 3 −1 5 + 5 −9 5 = 3i 5 + 15i 5 = 18i 5
16. 2i(3i2 − 4i + 7) = 6i3 − 8i2 + 14i = −6i + 8 + 14i = 8 + 8i
17. To find the conjugate, we change the addition sign to a subtraction sign. The conjugate of 11 + 13i is 11 − 13i.
18. (6 − 7i)(6 + 7i) = 36 + 42i − 42i − 49i2 = 36 + 49 = 85.
19. Since 8 + 4i = a + bi, a = 8 and bi = 4i, so b = 4.
20. Since 28i = a + bi, a = 0 and bi = 28i, so b = 28.
21. Since 6 = 3a + 5bi, 3a = 6 and 5bi = 0i. Thus, a = 2 and b = 0.
22. Since 15 − 25i = 3a + 5bi, 3a = 15 and 5bi = −25i, so 5b = −25. Thus, a = 5 and b = −5.
23. Since 36 + 12i = 9a − 3bi, 9a = 36 and −3bi = 12i, so −3b = 12. Thus, a = 4 and b = −4.
PROBLEMS
24. (a) i14 = i12 · i2 = (i4 )3 · i2 = 1 · −1 = −1
i24 = (i4 )6 = 1
(b) i30 = i28 · i2 = (i4 )7 · i2 = 1 · −1 = −1
i40 = (i4 )10 = 1
32
(c) i = (i4 )8 = 1
i48 = (i4 )12 = 1
19
(d) i = i16 · i3 = (i4 )4 · i3 = 1 · −i = −i
= i27 = i24 · i3 = (i4 )6 · i3 = 1 · −i = −i
Statements (c) and (d) are both true.
25. To solve this equation, we will use the quadratic formula.
x=
=
=
=
=
8±
√
64 − 68
√2
8 ± −4
2
8 ± 2i
2
2i
8
±
2
2
4 ± i.
9.5 SOLUTIONS
26. To solve this equation, we first set it equal to zero.
x2 + 29 = 10x
x2 − 10x + 29 = 0
Applying the quadratic formula gives
x=
=
=
=
=
√
100 − 116
√2
10 ± −16
2
10 ± 4i
2
10
4i
±
2
2
5 ± 2i.
10 ±
27. To solve this equation, we will use the quadratic formula.
x=
=
=
=
=
=
6±
√
36 − 56
√2
6 ± −20
2
√
√
6 ± −4 · 5
2
√
6 ± 2i 5
2 √
6
2i 5
±
2 √2
3 ± i 5.
28. To solve this equation, we first set it equal to zero.
x2 − 12x + 45 = 6
x2 − 12x + 39 = 0
Applying the quadratic formula gives
x=
=
=
=
=
=
√
144 − 156
√2
12 ± −12
2√
√
12 ± −4 · 3
2√
12 ± 2i 3
2 √
2i 3
12
±
2 √ 2
6 ± i 3.
12 ±
29. To solve this equation, we first set it equal to zero.
2x2 − 10x + 20 = 7
2x2 − 10x + 13 = 0
Applying the quadratic formula gives
x=
10 ±
√
100 − 104
4
341
342
Chapter Nine /SOLUTIONS
√
10 ± −4
4
10 ± 2i
=
4
10
2i
=
±
4
4
5
1
= ± i.
2
2
=
30. To solve this equation, we first set it equal to zero.
x2 − 5x + 16 = 3
x2 − 5x + 13 = 0
Applying the quadratic formula gives
x=
=
=
=
=
=
√
25 − 52
√2
5 ± −27
2
√
√
5 ± −9 · 3
2
√
5 ± 3i 3
2 √
3i 3
5
±
2
2
√
3 3
5
±
i.
2
2
5±
31. To solve this equation, we first set it equal to zero.
4x2 + 2x + 1 = 4x
4x2 − 2x + 1 = 0
Applying the quadratic formula gives
x=
=
=
=
=
=
2±
√
4 − 16
√8
2 ± −12
8
√
√
2 ± −4 · 3
8
√
2 ± 2i 3
8 √
2
2i 3
±
8
√8
1
3
±
i.
4
4
32. To solve this equation, we first set it equal to zero.
3x2 + 3x + 10 = 7x + 6
3x2 − 4x + 4 = 0
Applying the quadratic formula gives
x=
4±
√
16 − 48
6
SOLUTIONS to Review Problems for Chapter Nine
=
4±
4±
√
6
√
−32
−16 ·
√6
4 ± 4i 2
=
6 √
4i 2
4
= ±
6
√6
2
2 2
= ±
i.
3
3
=
√
343
2
Solutions for Chapter 9 Review
EXERCISES
1. The parabola opens up because the coefficient of x2 is 1, which is positive. The vertex is at (2, 0). Set x = 0 to find it has
a y-intercept of 4. See Figure 9.5.
y
9
7
5
3
1
x
1
−1
2
3
4
5
Figure 9.5
2. The parabola opens up because the coefficient of x2 is 2, which is positive. The vertex is at (−1, 5). Set x = 0 to find it
has a y-intercept of 7. See Figure 9.6.
y
13
11
9
7
5
3
1
−3
−2
−1
Figure 9.6
x
1
344
Chapter Nine /SOLUTIONS
3. The parabola opens down because the coefficient of x2 is −1, which is negative. The x-intercepts are at x = 3 and
x = −2. Set x = 0 to find it has a y-intercept of 6. See Figure 9.7.
y
6
3
x
−3
1
−1
3
−3
−6
Figure 9.7
4. The parabola opens up because the coefficient of x2 is 2, which is positive. The x-intercepts are at x = −3/2 and x = −3.
Set x = 0 to find it has a y-intercept of 9. See Figure 9.8.
y
9
6
3
−2
x
−4
−1
Figure 9.8
5. The factored form of the function tells us the x-intercepts are x = 1 and x = 5. Since the coefficient of x2 is positive, the
parabola opens up. Set x = 0 to find it has a y-intercept of 5. See Figure 9.9.
y
5
x
1
5
Figure 9.9
6. The factored form of the function tells us the x-intercepts are x = −3 and x = 4. Since the coefficient of x2 is negative,
the parabola opens down. Set x = 0 to find it has a y-intercept of 24. See Figure 9.10.
SOLUTIONS to Review Problems for Chapter Nine
345
y
24
4
−3
x
Figure 9.10
7. The vertex form of the function enables us to locate the vertex of the parabola. In this case, the vertex is at the point (3, 5).
Since the coefficient of x2 is positive, the parabola opens up. Set x = 0 to find it has a y-intercept of 23. See Figure 9.11.
y
23
(3, 5)
x
Figure 9.11
8. The vertex form of the function enables us to locate the vertex of the parabola. In this case, the vertex is at the point
(−1, 25). Since the coefficient of x2 is negative, the parabola opens down. Set x = 0 to find its y-intercept at 24. See
Figure 9.12.
(−1, 25)
y
24
x
Figure 9.12
9. This is already in standard form −5x2 − 2x + 3.
10. We have
7−
t2
t
1
1
− = 7 − t2 − t
2
3
2
3
1
1
= − t2 − t + 7.
2
3
11. We have
z 2 + 4z + 7
1 2
z + 4z + 7
=
5
5
4
7
1
= z2 + z + .
5
5
5
346
Chapter Nine /SOLUTIONS
12. We have
2(r − 2)(3 − 2r) = (2r − 4)(3 − 2r)
= 6r − 4r 2 − 12 + 8r
= −4r 2 + 14r − 12.
13. We have
2 (s − s(4 − s) − 1) = 2(s − (4s − s2 ) − 1)
= 2(s − 4s + s2 − 1)
= 2(−3s + s2 − 1)
= −6s + 2s2 − 2
= 2s2 − 6s − 2.
14. We have
(z 2 + 3)(z 2 + 2) − (z 2 + 4)(z 2 − 1) = z 4 + 2z 2 + 3z 2 + 6 − (z 4 − z 2 + 4z 2 − 4)
= z 4 + 2z 2 + 3z 2 + 6 − z 4 − 3z 2 + 4
= 2z 2 + 0 · z + 10.
15. We see that (x + 7)2 is a positive number or 0 for all values of x, that is, (x + 7)2 ≥ 0. Since its smallest value is 0, we
add that to −8 to get a minimum value of −8.
16. We see that (x + 2)2 is a positive number or 0 for all values of x, so, −(x + 2)2 ≤ 0. Since its largest value is 0, we add
that to a to get a maximum value of a.
17. We see that 7(x + a)2 is a positive number or 0 for all values of x, so −7(x + a)2 ≤ 0. Since its largest value is 0, we
add that to q to get a maximum value of q.
18. We rewrite this as 2(x + 3)2 + 2 (because x2 + 6x + 9 = (x + 3)2 ). We see that 2(x + 3)2 is a positive number or 0 for
all values of x, that is, 2(x + 3)2 ≥ 0. Since its smallest value is 0, we add that to 2 to get a minimum value of 2.
19. Since x2 − 6x + 9 = (x − 3)2 , we can rewrite this expression as
2(x2 − 6x + 9) + 4 = 2(x − 3)2 + 4,
so a = 2, h = 3, k = 4.
20. Since
(3 − x)2 = (−(x − 3))2 = (−1)2 (x − 3)2 = (x − 3)2 ,
we can rewrite this expression as
11 − 7(3 − x)2 = −7(x − 3)2 + 11.
We see that a = −7, h = 3, k = 11.
21. Rewriting this expression as
(2x + 4)2 − 7 = (2(x + 2))2 − 7
= 4(x + 2)2 − 7
factoring out 2
simplifying
2
= 4(x − (−2)) − 7,
we see that a = 4, h = −2, k = −7.
22. The vertex form looks like a(x − h)2 + k. We note that
x2 − 12x + 36 = (x − 6)2
by factoring, so (x − 6)2 is the vertex form (a = 1, h = 6, k = 0).
SOLUTIONS to Review Problems for Chapter Nine
23. The vertex form looks like a(x − h)2 + k. We note that
−x2 + 2bx − b2 = −(x2 − 2bx + b2 ) = −(x − b)2
by factoring, so −(x − b)2 is the vertex form (a = −1, h = b, k = 0).
24. The vertex form looks like a(x − h)2 + k. We note that
6(x2 − 8x + 16) + 2 = 6(x − 4)2 + 2
by factoring, so 6(x − 4)2 + 2 is the vertex form (a = 6, h = 4, k = 2).
25. We have
(t − 6)2 − 3
1
3
= (t − 6)2 − .
4
4
4
This is a quadratic expression in vertex form with a = 1/4, h = 6, k = −3/4.
26. The vertex form looks like a(x − h)2 + k. We note that
b + c(x2 − 4dx + 4d2 ) + 5 = c(x − 2d)2 + b + 5 = c(x − 2d)2 + (b + 5)
by factoring, so c(x − 2d)2 + (b + 5) is the vertex form (a = c, h = 2d, k = b + 5).
27. We factor the expression on the left.
x2 + 3x + 2 = 0
(x + 2)(x + 1) = 0.
The left-hand side is zero when x + 2 = 0 or x + 1 = 0, so the solutions are x = −1 and x = −2.
28. We rearrange the terms on the left, then factor:
1 + x2 + 2x = 0
x2 + 2x + 1 = 0
(x + 1)2 = 0.
The left-hand side is zero when x + 1 = 0, so there is one solution at x = −1.
29. We rearrange the terms on the left, then factor:
5z + 6z 2 + 1 = 0
6z 2 + 5z + 1 = 0
(3z + 1)(2z + 1) = 0.
The left-hand side is zero when 3z + 1 = 0 and when 2z + 1 = 0. Thus the solutions are z = − 31 and z = − 21 .
30. (a) Standard form:
y − 8 = −2(x + 3)2
y − 8 = −2(x2 + 6x + 9)
y − 8 = −2x2 − 12x − 18
y = −2x2 − 12x − 10,
so a = −2, b = −12, c = −10.
(b) From part (a), we know that y = −2x2 − 12x − 10. We have
y = −2x2 − 12x − 10
= −2(x2 + 6x + 5)
= −2(x + 5)(x + 1)
= −2(x − (−5))(x − (−1)),
so a = −2, r = −5, s = −1.
347
348
Chapter Nine /SOLUTIONS
(c) This is already almost in vertex form. We have
y − 8 = −2(x + 3)2
y = −2(x − (−3))2 + 8,
so that a = −2, h = −3, k = 8
31. (a) Multiplying out, we have
y = 2x(3x − 7) + 5(7 − 3x)
= 6x2 − 14x + 35 − 15x
= 6x2 − 29x + 35,
so a = 6, b = −29, c = 35.
(b) This is already almost in factored form. We have
y = 2x(3x − 7) + 5(7 − 3x)
= 2x(3x − 7) + 5(−1)(3x − 7)
factor out −1
= (3x
−5) 5 − 7)(2x
7
·2 x−
= 3 x−
3 2
5
7
x−
,
= 6 x−
3
2
factor out (3x − 7)
= 2x(3x − 7) − 5(3x − 7)
factor out coefficients
so a = 6, r = 7/3, s = 5/2.
(c) To complete the square, we begin with standard form from part (a):
y = 6x2 − 29x + 35
= 6x2 − 29x
29
= x2 −
x
6
2
29
29
2
=x −
x+
6
12
29 2 841
−
= x−
12 2 144 29
841
−
y = 6 x−
+ 35
12
144
2
29
841
840
y = 6 x−
−
+
12 2
24
24
1
29
−
,
y = 6 x−
12
24
y − 35
y − 35
6 2
29
y − 35
+
6
12
y − 35
6
so a = 6, h = 29/12, k = −1/24.
32. We rewrite the equation as
x2 = 16,
and take the square root of each side, getting
√
x = ± 16 = ±4.
33. We subtract 5 from both sides
x2 = 4,
and take the square root of each side, getting
√
x = ± 4 = ±2.
because
1
2
·
29
6
=
29
12
SOLUTIONS to Review Problems for Chapter Nine
34. We take the square root of each side, getting
√
x + 4 = ± 25 = ±5,
and subtract 4 from both sides
x + 4 = ±5
x = ±5 − 4
x = 1 and − 9.
35. We subtract 7 from each side:
(x − 6)2 = 0.
We take the square root of each side, getting
x − 6 = 0.
We add 6 to both sides:
x = 6.
36. We subtract 15 from each side:
We take the square root of each side, getting
We add 5 to both sides:
(x − 5)2 = 2.
√
x − 5 = ± 2.
√
x = ± 2 + 5.
37. The equation is already in factored form, so x = 1 and x = 2.
38. The equation is already in factored form, so x = 93 and x = −115.
39. The equation is already in factored form, so x = −47 and x = −59.
40. The left side of this equation can be factored, giving:
x2 + 14x + 45 = 0
(x + 9)(x + 5) = 0.
Thus x = −9 and x = −5.
41. The left side of this equation can be factored, giving:
x2 + 21x + 98 = 0
(x + 14)(x + 7) = 0.
Thus x = −14 and x = −7.
42. The left side of this equation can be factored, giving:
2x2 + x − 55 = 0
(2x + 11)(x − 5) = 0.
Thus 2x + 11 = 0 and x − 5 = 0, giving
2x + 11 = 0
2x = −11
11
x=− ,
2
349
350
Chapter Nine /SOLUTIONS
and
x−5 = 0
x = 5.
43. The left side of this equation can be factored, giving:
x2 + 16x + 64 = 0
(x + 8)(x + 8) = 0.
Thus x = −8.
44. The left side of this equation can be factored, giving:
x2 + 26x + 169 = 0
(x + 13)(x + 13) = 0.
Thus x = −13.
45. Use the quadratic formula or complete the square as follows:
Step 1. Add −7 to both sides, giving x2 + 12x = −7.
Step 2. Since 12/2 = 6 and 62 = 36, we add 36 to both sides:
x2 + 12x + 36 = −7 + 36
Step 3. Rewriting the left side and combining terms on the right side gives
(x + 6)2 = 29.
Step 4. Taking square roots, we get
(x + 6) =
√
so
x = −6 +
29
√
29
and
and
√
(x + 6) = − 29,
x = −6 −
√
29.
46. Use the quadratic formula or complete the square as follows:
Step 1. Add 56 to both sides, giving x2 + 24x = 56.
Step 2. Since 24/2 = 12 and 122 = 144, we add 144 to both sides:
x2 + 24x + 144 = 56 + 144.
Step 3. Rewriting the left side and combining terms on the right side gives
(x + 12)2 = 200.
Step 4. Taking square roots, we get
(x + 12) =
√
so
x = −12 +
200
√
200
and
and
√
(x + 12) = − 200,
x = −12 −
√
200.
SOLUTIONS to Review Problems for Chapter Nine
351
47. We first divide both sides by 3, giving x2 + 10x + 3 = 0.
Use the quadratic formula or complete the square as follows:
Step 1. Add −3 to both sides, giving x2 + 10x = −3.
Step 2. Since 10/2 = 5 and 52 = 25, we add 25 to both sides:
x2 + 10x + 25 = −3 + 25.
Step 3. Rewriting the left side and combining terms on the right side gives
(x + 5)2 = 22.
Step 4. Taking square roots, we get
(x + 5) =
√
so
x = −5 +
22
√
√
(x + 5) = − 22,
and
22
x = −5 −
and
√
22.
48. Factoring out a 2x we get
2x(x2 − 2x + 1) = 0
2x(x − 1)2 = 0,
giving x = 0 and x = 1.
49. Factoring out an x we get x(x4 + 3x2 + 2) = 0. The factor inside the parentheses can be regarded as a quadratic in x2 .
So we get
x((x2 )2 + 3(x2 ) + 2) = 0
x(x2 + 1)(x2 + 2) = 0.
The factors x2 + 1 and x2 + 2 are never zero, so the only solution is x = 0.
√
50. If we let y = x, the equation becomes y 2 + y = 6, or y 2 + y − 6 = 0. This
√is a quadratic in y which can be factored as
(y − 2)(y + 3) = 0, so it has solutions y = 2 and y = −3. However, y = x ≥ 0, so y = 2 is the only valid solution.
Because x = y 2 , we have x = 4.
51. If we multiply x−1 + 2−1 = −x − 2 by 2x, the equation becomes 2 + x = −2x2 − 4x, or 2x2 + 5x + 2 = 0. This is a
quadratic equation in x which can be factored, (2x + 1)(x + 2) = 0, with solutions x = −1/2 and x = −2.
52. We have a = 1, b = −4, c = −12, so
x=
4±
p
(−4)2 − 4(1)(−12)
4±
=
2(1)
√
√
4 ± 64
4±8
16 + 48
=
=
.
2
2
2
This gives x = 12/2 = 6 and x = −4/2 = −2.
53. We have a = 2, b = −5, c = −12, so
x=
5±
p
(−5)2 − 4(2)(−12)
5±
=
2(2)
√
√
25 + 96
5 ± 121
5 ± 11
=
=
.
4
4
4
This gives x = 16/4 = 4 and x = −6/4 = −3/2.
54. To have 0 on the right side of this equation, we subtract 6 from both sides:
y 2 + 3y − 2 = 0.
Thus, a = 1, b = 3, c = −2, so
y=
−3 ±
√
√
(3)2 − 4(1)(−2)
−3 ± 9 + 8
−3 ± 17
=
=
.
2(1)
2
2
p
352
Chapter Nine /SOLUTIONS
55. To have 0 on the right side of this equation, we subtract 7 from both sides:
3y 2 + y − 9 = 0.
Thus, a = 3, b = 1, c = −9, so
y=
−1 ±
√
√
(1)2 − 4(3)(−9)
−1 ± 1 + 108
−1 ± 109
=
=
.
2(3)
6
6
p
56. To have 0 on the right side of this equation, we subtract 3 from both sides:
7t2 − 15t + 2 = 0.
Thus a = 7, b = −15, c = 2, so
t=
15 ±
√
√
(−15)2 − 4(7)(2)
15 ± 225 − 56
15 ± 169
15 ± 13
=
=
=
.
2(7)
14
14
14
p
This gives t = 28/14 = 2 and t = 2/14 = 1/7.
57. To have 0 on the right side of this equation, we subtract 2t + 8 from both sides:
−2t2 + 10t − 3 = 0.
Thus, a = −2, b = 10, c = −3, so
√
√
(10)2 − 4(−2)(−3)
−10 ± 100 − 24
−10 ± 76
=
=
.
t=
2(−2)
−4
−4
√
√
√
We can rewrite 76 as 4 · 19 = 2 19. Thus,
√
−10 ± 2 19
t=
.
−4
√
√
√
√
The two solutions are (−10 + 2 19)/(−4) = (5 − 19)/2 and (−10 − 2 19)/(−4) = (5 + 19)/2.
−10 ±
p
58. Step 1. Add −8 to both sides, giving x2 − 8x = −8.
Step 2. Since−8/2 = −4 and (−4)2 = 16, we add 16 to both sides:
x2 − 8x + 16 = −8 + 16.
Step 3. Rewriting the left side gives
Step 4. Taking square roots, we get
so
(x − 4)2 = 8.
√
x − 4 = ± 8,
x=4±
√
8.
2
59. Step 1. Add 2 to both sides, giving y + 10y = 2.
Step 2. Since 10/2 = 5 and 52 = 25, we add 25 to both sides:
y 2 + 10y + 25 = 2 + 25.
Step 3. Rewriting the left side gives
(y + 5)2 = 27.
Step 4. Taking square roots, we get
so
√
y + 5 = ± 27,
y = −5 ±
√
27.
SOLUTIONS to Review Problems for Chapter Nine
60. Step 1. Add 1 to both sides, giving s2 + 3s = 3.
Step 2. Since (3/2)2 = 9/4, we add 9/4 to both sides:
s2 + 3s +
9
9
=3+ .
4
4
Step 3. Rewriting the left side gives
(s +
3 2
21
) =
.
2
4
Step 4. Taking square roots, we get
3
=±
2
r
21
,
4
3
s=− ±
2
r
21
.
4
s+
so
√
−3 ± 21
Simplifying, we get s =
.
2
61. Step 1. Add −2 to both sides, giving r 2 − r = 5.
Step 2. Since (−1/2)2 = 1/4, we add 1/4 to both sides:
r2 − r +
1
1
=5+ .
4
4
Step 3. Rewriting the left side gives
(r −
1 2
21
) =
.
2
4
Step 4. Taking square roots, we get
1
r− =±
2
so
1±
1
r= ±
2
√
r
r
21
,
4
21
.
4
21
.
2
62. Since a = 2, we first divide by 2, giving t2 − 2t + 2 = 3.
Step 1. Add −2 to both sides, giving t2 − 2t = 1.
Step 2. Since −2/2 = −1 and (−1)2 = 1, we add 1 to both sides:
Simplifying, we get r =
t2 − 2t + 1 = 1 + 1.
Step 3. Rewriting the left side gives
(t − 1)2 = 2.
Step 4. Taking square roots, we get
√
t − 1 = ± 2,
so
t=1±
√
2.
2
63. Since a = 3, we first divide by 3, giving v + 3v = 4.
Step 1. Since 3/2 = 23 and ( 23 )2 = 94 , we add 49 to both sides:
v 2 + 3v +
Step 2. Rewriting the left side gives
(v +
9
9
=4+ .
4
4
3 2
25
) =
.
2
4
353
354
Chapter Nine /SOLUTIONS
Step 3. Taking square roots, we get
3
v+ =±
2
so
r
3
v=− ±
2
r
25
,
4
−3 ± 5
25
=
.
4
2
We get two solutions, v = 1 and v = −4.
64. The expression (x − 3)2 is always positive or zero, so the left-hand side of this equation is always greater than or equal
to zero. The right-hand side of this equation is negative, so the two sides can never be equal.
65. If x is positive, then the expression x2 + x is also positive and the expression x2 + x + 7 is also positive and greater than
7. Therefore, for positive x, the left-hand side can never be zero, and so the equation has no positive solution.
66. In factored form, we have
(x − 2)(x − (−6)) = (x − 2)(x + 6)
= x2 + 4x − 12.
67. In factored form, we have
t−
2
3
t−
t− −
2
3
1
3
t+
1
3
=0
= 0.
Multiplying both sides by 3, and then 3 again, gives
2
1
·3 t+
= 0·9
3
3
(3t − 2)(3t + 1) = 0
3 t−
9t2 − 3t − 2 = 0.
68. In factored form, we have
√ √ x − (2 + 5) x − (2 − 5) = 0
√
√
√
√
x2 − x(2 − 5) − x(2 + 5) + (2 + 5)(2 − 5) = 0
√
√
√
√
√
x2 − 2x + x 5 − 2x − x 5 + 4 − 2 5 + 2 5 − ( 5)2 = 0
x2 − 4x − 1 = 0.
If we recognize that this solution looks like it comes from the vertex form, we have
√
0 = (x − h)2 + k
± k = x−h
x = −h ±
so h = 2 and k = −5, giving
√
k,
(x − 2)2 − 5 = x2 − 4x + 4 − 5 = x2 − 4x − 1,
which is the same as we found with the factored form.
69. In factored form, we have
√
√
(x − 5)(x + 3) = 0
√
√
√
√
x2 + x 3 − x 5 − 5 3 = 0
√
√
√
x2 + ( 3 − 5)x − 15 = 0.
SOLUTIONS to Review Problems for Chapter Nine
355
70. (10 + 3i) − (18 − 4i) = (10 + 3i) + (−18 + 4i) = (10 − 18) + (3i + 4i) = −8 + 7i.
71.
72.
73.
74.
75.
76.
4−i 3−i
12 − 4i − 3i + i2
12 − 7i − 1
11 − 7i
11
7
4−i
=
·
=
=
=
=
−
i
3+i
3+i 3−i
9 − i2
9+1
10
10
10
√
√
√
−100 = 100 −1 = 10i
√
√
−9 + 5 −49 = 3i + 5(7i) = 3i + 35i = 38i
√
√
√
√ √
√
√
√ √
−8 + 3 −18 = −4 2 + 3 −9 2 = 2i 2 + 9i 2 = 11i 2
√
√
√
√
√ √
√ √
√
2 −3 − 4 −27 = 2 −1 3 − 4 −9 3 = 2i 3 − 12i 3 = −10i 3
√
√
√
√
√ √
√
√
√
5 −32 − 6 −2 = 5 −16 2 − 6 −1 2 = 20i 2 − 6i 2 = 14i 2
77. (3 − i)(5 + 3i) = 15 + 9i − 5i − 3i2 = 15 + 4i + 3 = 18 + 4i
5i
2−i
10i − 5i2
5 + 10i
5
10
10i + 5
5i
=
·
=
=
= +
i = 1 + 2i
=
2+i
2+i 2−i
4 − i2
4+1
5
5
5
79. (1 + 4i) + (6 − 8i) = (1 + 6) + (4i − 8i) = 7 − 4i
78.
80. 4i3 (6i5 ) − 2i(4i8 ) + 3i2 = 24i8 − 8i9 + 3i2 = 24 − 8i − 3 = 21 − 8i
3
i2
7
1
31
7
i + 5i −
= 15 + i + =
+ i
2
2
2
2
2
2
82. (3i)3 + (5i)2 − 4i + 30 = 27i3 + 25i2 − 4i + 30 = −27i − 25 − 4i + 30 = 5 − 31i
81. (3 + i)(5 − i/2) = 15 −
83.
10 + 25i − 10
5 + 10i
5 + 10i 2 + i
10 + 5i + 20i + 10i2
0 + 25i
0
25
=
=
·
=
=
= +
i = 5i
2−i
2−i
2+i
4 − i2
4+1
5
5
5
PROBLEMS
84. (a) We show that the factored and vertex forms are equivalent to the standard form. For the factored form:
(x − 2)(x − 8) = x2 − 2x − 8x + 16
= x2 − 10x + 16.
For the vertex form:
(x − 5)2 − 9 = (x2 − 10x + 25) − 9
= x2 − 10x + 16.
(b)
In each case, we get the given standard form, so all three expressions are equivalent.
(i) The vertex form gives the smallest (or largest) value of the expression. In this case, since (x − 5)2 is always
positive or zero, the expression (x − 5)2 − 9 is always greater than or equal to −9. The smallest value of the
expression is −9.
(ii) The factored form gives the values of x that make the expression zero. Since factored form is (x − 2)(x − 8),
the expression is zero when x = 2 or x = 8.
(iii) The standard form is the easiest one to use to find the value of the expression when x = 0. Substituting x = 0,
we see that 02 − 10 · 0 + 16 = 16. The value of the expression when x = 0 is the constant term 16.
85. (a) Since the parabola opens up, we know that a is positive. Since c gives the value of the function when x = 0, we see
that c is the y-intercept, so c = 21.
(b) The parameters r and s in factored form are the values of x that make the expression zero, so they are the x-intercepts.
We have r = 3 and s = 7.
(c) The vertex of the parabola is at the point (h, k). Since the vertex of this parabola is at (5, −4), we have h = 5 and
k = −4.
86. (a) Since the parabola opens down, we know that a is negative. Since c gives the value of the function when x = 0, we
know that c is the y-intercept, so c = 12.
(b) The parameters r and s in factored form are the values of x that make the expression zero, so they are the x-intercepts.
We have r = −2 and s = 6.
(c) The vertex of the parabola is at the point (h, k). Since the vertex of this parabola is at (2, 16), we have h = 2 and
k = 16.
356
Chapter Nine /SOLUTIONS
87. We have
y = 4x − 30 + 2x2
= 2x2 + 4x + (−30),
so a = 2, b = 4, c = −30.
88. We have
y = 4x − 30 + 2x2
= 2x2 + 4x − 30
= 2(x2 + 2x − 15)
= 2(x + 5)(x − 3)
= 2(x − (−5))(x − 3),
so a = 2, r = −5, s = 3.
89. We have
y = 4x − 30 + 2x2
y + 30
y + 30
2
y + 30
+1
2
y + 30
+1
2
y + 30
2
y + 30
= 2x2 + 4x
= x2 + 2x
= x2 + 2x + 1
= (x + 1)2
= (x + 1)2 − 1
= 2(x + 1)2 − 2
y = 2(x + 1)2 − 32
y = 2(x − (−1))2 − 32,
so a = 2, h = −1, k = −32.
90. We have
y = 4x − 30 + 2x2
= 2x2 + 4x − 30
= 2x(x + 2) − 30
= 2x(x − (−2)) + (−30),
so a = 2, v = −2, w = −30.
91. Since the term 3(x − 6)2 is always positive or zero, we see that the expression for f is in the form
f (x) = (Positive or zero term) + 10.
Thus the values of the function are greater than or equal to 10 for all x. The smallest value is 10, and it occurs when
3(x − 6)2 is zero, which is when x = 6.
92. Since the term −4(t + 1)2 is always negative or zero, we see that the function is in the form
g(t) = (Negative or zero term) + 21.
Thus the values of the function are less than or equal to 21 for all t. The largest value is 21, and it occurs when −4(t + 1)2
is zero, which is when t = −1.
SOLUTIONS to Review Problems for Chapter Nine
357
93. Factoring we get x2 − 6x + 9 = (x − 3)2 . Notice that any real number squared must be greater than or equal to zero, so
(x − 3)2 ≥ 0.
94. (a) We have
y = x2 − 6x + 20
y − 20 = x2 − 6x
y − 20 + 9 = x2 − 6x + 9
y − 11 = (x − 3)2
y = (x − 3)2 + 11.
The vertex form of the quadratic expression x2 − 6x + 20 is (x − 3)2 + 11.
(b) The smallest value of the function is 11, and it takes this value when x = 3.
95. (a) The stone is at the top of the towers at time t = 0. The height of the towers is given by
h = 1483 − 16 · 02 = 1483 feet.
(b) The stone hits the ground when h = 0, so
1483 − 16t2 = 0
16t2 = 1483
t=±
r
1483
= ±9.627 seconds.
16
Since the stone is dropped at time t = 0, the positive solution, 9.627 seconds, represents the time at which it hits the
ground.
96. We need to find out when the height of the rocket is zero. We can factor 16t from the expression −16t2 + 160t to give us
16t(−t + 10). This expression is zero when t = 0 (the time the rocket is launched) and when t = 10. Thus, the rocket is
in the air for 10 seconds.
97. (a) If we denote the opposite corner by (x, y), then the area of the rectangle is A = xy. However, y = −x + b, so the
area is A = xy = x(−x + b) = −x2 + bx.
(b) The graph of the function A = −x2 + bx is a parabola opening down so it attains its maximum at the vertex. Thus,
we rewrite A = −x2 + bx in vertex form, A = −x2 + bx = −(x − b/2)2 + b2 /4. The maximum occurs when
x = b/2.
(c) When x = b/2 we find y = −x + b = −b/2 + b = b/2. Thus, x = y = b/2, so the rectangle is a square.
98. (a) The 200 feet of fencing would be the perimeter of the rectangular region formed. If the paddock has length l and
width w, then 200 = 2l + 2w or l + w = 100. We can solve for one variable in terms on another: l = 100 − w.
Since A = lw, we now have A = (100 − w)w.
(b) The maximum area is determined by finding the vertex of the parabola.
A = (100 − w)(w)
A = 100w − w2
−A = w2 − 100w
−A + 502 = w2 − 100w + 502
−A + 502 = (w − 50)2
−A = (w − 50)2 − 502
A = −(w − 50)2 + 502
A = −(w − 50)2 + 2500.
The vertex of the parabola is (50, 2500). Thus, the maximum area is 2500 square feet.
358
Chapter Nine /SOLUTIONS
99. (a) The 200 feet of fencing would be the perimeter of the rectangular region formed plus the length of the extra piece of
fencing. Let y be the length in feet of the side the paddocks share, and let w be the total width of both paddocks. Since
there are three sides of length y and two of length w, we have 200 = 2w + 3y. We can solve for one variable in terms
on another: w = 0.5(200 − 3y). Since the total area of both paddocks is wy, we have A = wy = 0.5(200 − 3y)y =
(200 − 3y)y/2.
(b) The maximum area is determined by finding the vertex of the parabola.
1
(200 − 3y)(y)
2
2A = 200y − 3y 2
2
200
− A = y2 −
y
3
3
2
200
100
100 2
2
= y2 −
y+
− A+
3
3
3
3
2
100 2
100 2
− A+
= y−
3
3
3
100 2
100 2
2
− A= y−
−
3
3
3
100 2 3 100 2
3
y−
+
A=−
2
3
2
3
100 2 3 10,000
3
y−
+
=−
2
3
2
9
3
100 2 10,000
=−
y−
.
+
2
3
6
100 10,000
10,000
The vertex of the parabola is
. Thus, the maximum area is
,
= 1666.667 square feet.
3
6
6
100. The total length around the outside of the plot is x + x + x + x = 4x, so the cost of the hedges is 10 · 4x = 40x dollars.
The area of the inside of the plot is x · x = x2 so the cost of the seed is 0.01x2 dollars. The total cost is 0.01x2 + 40x
dollars.
A=
101. This is in factored form, and we see that it has two x-intercepts, at x = −3 and at x = 5. See Figure 9.13.
y
y = −3(x + 3)(x − 5)
x
5
−3
Figure 9.13
102. This is in factored form with y = 1.5(x − 10)(x − 10). There is a single x-intercept, at x = 10. This point is also the
vertex of the parabola. See Figure 9.14.
y
y = 1.5(x − 10)2
x
10
Figure 9.14
SOLUTIONS to Review Problems for Chapter Nine
359
103. The vertex of this parabola is at the point (1, 5), in the first quadrant. Since a is negative, the parabola opens down, and it
must have two x-intercepts. See Figure 9.15.
y
(1, 5)
y = −(x − 1)2 + 5
x
Figure 9.15
104. The vertex of this parabola is at the point (−3, 4), in the second quadrant. Since a is positive, the parabola opens up. Since
the vertex is above the x-axis, and the parabola opens up, it has no x-intercepts. See Figure 9.16.
y
y = 2(x + 3)2 + 4
(−3, 4)
x
Figure 9.16
105. We observe the horizontal line y = 8. Multiplying the expression out, we find
y = (x + 2)(x + 1) + (x − 2)(x − 3) − 2x(x − 1) = (x2 + 3x + 2) + (x2 − 5x + 6) − 2x2 + 2x = 8.
106. We observe that (x − 1)(x − 2) + 3x and x2 are identical except (x − 1)(x − 2) + 3x has been moved 2 units up the
y-axis. Multiplying the expression out, we find
(x − 1)(x − 2) + 3x = x2 − 3x + 2 + 3x = x2 + 2.
Thus the values of y = (x − 1)(x − 2) + 3x are always 2 more than the values of y = x2 . This explains why the graph
is 2 units higher.
107. All three graphs are the same, namely y = x2 . Now,
y = (x − a)2 + 2a(x − a) + a2 = (x2 − 2ax + a2 ) + (2ax − 2a2 ) + a2 = x2 .
108. Let x represent the number.
x2 + x = 3x + 48
2
x2 − 2x = 48
x − 2x + 1 = 48 + 1
(x − 1)2 = 49
√
x − 1 = ± 49
x = 1±7
x = 8 or
x = −6.
360
Chapter Nine /SOLUTIONS
109. If the equation has a positive solution, then we can substitute a positive value of x to solve the equation. For any positive
value of x, the expression x2 +x is positive so for positive values of x, the left-hand side is of the form Positive number+A.
The only way we can have Positive number + A = 0 is if A is negative. This equation has a positive solution if A is
negative.
110. This equation has the two solutions of x = −4 and x = −A. The solution −A is positive if A is negative, so this equation
has a positive solution if A is negative.
111. Notice that the equation has no solution when A = 0. If A 6= 0, we begin to solve this equation by adding 6 to both sides
and dividing by A to obtain (x − 3)2 = 6/A. This equation has no solution if A is negative. If A is positive, we have
a positive solution when we consider the positive square root and add 3. This equation has a positive solution when A is
positive.
112. We begin to solve this equation by adding A to both sides and dividing by 5 to obtain (x − 3)2 = A/5. This equation
has no solution if A is negative. If A = 0, then the equation has the positive solution x = 3. If A is positive, we have
a positive solution when we consider the positive square root and add 3. This equation has a positive solution when A is
positive or zero.
113. Substituting the solution x = 0 gives c = 0 so the only condition is that c = 0. There are no restrictions on a or b.
114. The factored form of the left-hand expression gives us the two solutions x = r and x = s. We have x = 0 as a solution if
r = 0 or s = 0.
115. Substituting x = 0 in this equation gives ah2 + k = 0. All three parameters must be related in this way if x = 0 is a
solution. The conditions on the parameters are that together they solve the equation ah2 + k = 0.
116. The solutions occur when (x − a)(x − 1) = 0, that is x = a and x = 1.
117. The solutions occur when (x − a)(x − 1) = 0. Since x 6= 1 we can divide by sides by x − 1 and get x − a = 0 so the
solution is a = x.
118. (x − a)(x − 1) = 0 has solutions x = 1 and x = a. There will be just one solution if and only if a = 1.
119. If x = 1 then x − 1 = 0 and the left-hand side is zero no matter what the value of a, so any value of a is a solution.
120. The 10th power of a number is always greater than or equal to zero, so there are no solutions if a > 0, since in that case
the left-hand side of the equation would be greater than 0. If a ≤ 0 there are solutions, because then −a ≥ 0, so we can
say 3x − 2 = (−a)1/10 and solve for x.
121. All values except a = −6. If a = −6 then the equation reduces to −5 = 0, which is impossible. However, if a 6= −6
then the equation is (a + 6)x − 5 = 0 which is linear in x and has a solution.
122. a = −5. The x terms cancel, reducing this equation to a = −5. This only has solutions if a = −5, and in that case, any
value of x is a solution.
123. This has solutions for all values of a. When you expand (1 − 2x)2 you get 1 − 4x + 4x2 , so the x2 terms cancel and this
leaves you with a linear equation in x.
124. This equation is equivalent to (a − 1)x2 + x = 0, which factors as x((a − 1)x + 1) = 0. Thus x = 0 is always a solution,
and when a = 1 it is the only one. When a 6= 1, then x = −1/(a − 1) is also a solution, different from 0. Thus there is
exactly one solution only when a = 1.
125. (a) We can rewrite (x − h)2 + k = 0 as (x − h)2 = −k. For this to have two distinct solutions, −k must be positive, so
k must be negative.
(b) The factored form of the right-hand side (x − r)(x − s) makes it clear that the right-hand side has two solutions when
it is set equal to zero. Since the left-hand side is equal to the right-hand side, it also must have two solutions when it
is set equal to zero. So by part (a), k must be negative.
(c) We use the same reasoning. When we set the right-hand side equal to zero, we arrive at the equation −2(x−h)2 = −k
which only has solutions if k is positive. We conclude that k must be positive.
126. (a) We know that the left-hand side expression is greater than or equal to k for all values of x since 2(x − h)2 is positive
or zero for all x.
(b) By part (a), the right-hand side expression is also greater than or equal to k for all x. When x = 0, the right-hand
side is c, so we see that c ≥ k.
(c) Since −2(x − h)2 is negative or zero for all x, we know that −2(x − h)2 + k is less than or equal to k for all x.
SOLUTIONS to Review Problems for Chapter Nine
361
Since −2(x − h)2 + k = ax2 + bx + c, it must also be true that ax2 + bx + c is less than or equal to k for all x. In
particular, for x = 0, we have c ≤ k.
127. We have
x2 + (a − 1)x − a = 0
x(x − 1) + a(x − 1) = 0
(x + a)(x − 1) = 0.
The solutions are x = −a and x = 1.
128. We rearrange the terms on the left then complete the square:
2a + 2 + a2 = 0
a2 + 2a + 2 = 0
(a + 1)2 + 1 = 0.
This has no solutions because the left-hand side is a square plus one. A square is always non-negative, so the left-hand
side is always positive.
Solutions for Solving Drill
1. This is a quadratic equation in x, so we have a variety of ways to solve it. This particular quadratic expression factors
easily so we solve it by factoring. We have
x2 − 4x + 3 = 0
(x − 3)(x − 1) = 0
x = 3 , x = 1.
2. This is a quadratic equation in x. We solve it by first putting all terms on one side so the expression is equal to zero. This
particular quadratic expression factors easily so we solve it by factoring. We have
2
10 + 3x = 18 + 5x − x2
x − 2x − 8 = 0
(x − 4)(x + 2) = 0
x = 4 , x = −2.
3. This is a linear equation in t. We solve it by putting terms with a t on one side and terms without a t on the other. We have
6 − 3t = 2t − 10 + 5t
6 − 3t = 7t − 10
16 = 10t
t = 1.6.
362
Chapter Nine /SOLUTIONS
4. This is a quadratic equation in t. Since we don’t easily see how to factor it, we choose to solve it using the quadratic
formula. We have
√
−b ± b2 − 4ac
t=
2a
p
−7 ± (7)2 − 4(1)(3)
=
2
√
−7 ± 37
=
.
2
The decimal approximations to these two solutions are t = −0.459 and t = −6.541.
5. This is a quadratic equation in r. We solve it by first putting all terms on one side so that the expression is set equal to
zero. We have
8r + 5 = r 2 + 5r + 4
2
r − 3r − 1 = 0.
The quadratic expression on the left does not easily factor, so we use the quadratic formula with a = 1, b = −3, c = −1:
√
−b ± b2 − 4ac
r=
2a p
−(−3) ± (−3)2 − 4(1)(−1)
=
2(1)
√
3 ± 13
=
.
2
The decimal approximations to these two solutions are r = −0.303 and r = 3.303.
6. We isolate the power s3 and raise both sides to the 1/3 power:
5s3 + 33 = 0
5s3 = −33
s3 = −6.6
s = (−6.6)1/3 = −1.876.
7. This is a quadratic equation in w. We solve it by first putting all terms on one side so the expression is equal to zero. This
particular quadratic expression factors and we solve it by factoring. We could also have used the quadratic formula if we
did not see how to factor it. We have
2w2 + 16w + 15 = 5w + 3
2w2 + 11w + 12 = 0
(2w + 3)(w + 4) = 0
w = −1.5 , w = −4.
8. This is a linear equation in x. We solve it by combining like terms and then putting all terms with x on one side of the
equation and all terms without x on the other. We have
5x + 13 − 2x = 10x − 9
3x + 13 = 10x − 9
−7x = −22
22
x=
.
7
SOLUTIONS to Review Problems for Chapter Nine
363
9. This is a quadratic equation in p. We solve it by first putting all terms on one side so the expression is equal to zero. We
have
3p2 + 3p − 8 = p2 + 5p − 14
2p2 − 2p + 6 = 0.
The quadratic expression on the left does not easily factor, so we use the quadratic formula.
√
−b ± b2 − 4ac
p=
p 2a
2 ± (−2)2 − 4(2)(6)
=
2(2)
√
√
√
2 ± −44
2 ± 2i 11
1 ± i 11
=
=
=
.
4
4
2
10. This is a quadratic equation in t. We solve it by first putting all terms on one side so the expression is equal to zero. Since
the resulting quadratic expression factors easily, we solve the equation by factoring. We have
3t2 + 5t − 4 = 2t2 + 8t + 6
t2 − 3t − 10 = 0
(t − 5)(t + 2) = 0.
t = 5 , t = −2.
11. Both sides of this equation are power functions. We solve it by first putting all terms on one side, so the expression is
equal to zero. We then solve the equation by factoring:
2x5 = 12x2
5
2x − 12x2 = 0
2x2 (x3 − 6) = 0,
Thus, by the zero-product rule, x2 = 0 or x3 = 6, so x = 0 or x = 61/3 = 1.817.
12. The expression on the left of the equal sign is quadratic and it does not easily factor. We solve this equation using the
quadratic formula. We have
√
−b ± b2 − 4ac
w=
2a
p
−5.2 ± (5.2)2 − 4(−2.3)(7.2)
=
2(−2.3)
√
−5.2 ± 93.28
.
=
−4.6
The decimal approximations for these two roots are w = −0.969 and w = 3.230.
13. Since this equation involves quadratic expressions, we begin to solve it by putting all terms on one side of the equation so
that it is set equal to zero. Since the resulting quadratic expression does not easily factor, we solve this equation using the
quadratic formula. We have
2.1q 2 − 4.2q + 9.8 = 14.9 + 1.7q − 0.4q 2
2.5q 2 − 5.9q − 5.1 = 0.
364
Chapter Nine /SOLUTIONS
Using the quadratic formula, we have
√
b2 − 4ac
2a
p
5.9 ± (−5.9)2 − 4(2.5)(−5.1)
=
2(2.5)
√
5.9 ± 85.81
.
=
5
q=
−b ±
The decimal approximations for these two roots are q = −0.673 and q = 3.033.
14. This equation is linear in t, so we begin to solve it by combining like terms and then putting all terms with t on one side
of the equation and all terms without t on the other. We have
4.2t + 5.9 − 0.7t = 12.2
3.5t + 5.9 = 12.2
3.5t = 6.3
t = 1.8.
15. This equation can be viewed as either an equation involving powers or an equation that is quadratic in x. Solving it using
either approach will work. If we solve it as a power, we first isolate the x2 and then take the square root. We have:
5x2 = 20
x2 = 4
x = ±2.
If we solve the equation instead by thinking of it as an equation that is quadratic in x, we first put all terms on one side of
the equation so that it is set equal to zero. The expression then can be factored to arrive at the solutions:
5x2 = 20
2
5x − 20 = 0
5(x2 − 4) = 0
5(x − 2)(x + 2) = 0
x = 2 , x = −2.
Notice that, as expected, we get the same answer no matter which approach we take.
16. We begin by using the distributive law. The resulting equation is quadratic in x, so we combine terms so that an expression
is set equal to zero. We have:
2x(x + 5) − 4x + 9 = 0
2x2 + 10x − 4x + 9 = 0
2x2 + 6x + 9 = 0.
Since the quadratic expression does not easily factor, we use the quadratic formula. We have
√
−b ± b2 − 4ac
x=
2a
p
−6 ± (6)2 − 4(2)(9)
=
2(2)
√
−6 ± −36
−6 ± 6i
3
3
=
=
= − ± i.
4
4
2
2
Since the square root of a negative number is not a real number, there are no real solutions to this equation.
SOLUTIONS to Review Problems for Chapter Nine
365
17. We begin by using the distributive law. The resulting equation is quadratic in p, so we combine terms and set the expression
equal to zero. We have:
p(p − 3) = 2(p + 4)
2
p2 − 3p = 2p + 8
p − 5p − 8 = 0.
Since the quadratic expression does not easily factor, we use the quadratic formula. We have
√
−b ± b2 − 4ac
p=
2a
p
−(−5) ± (−5)2 − 4(1)(−8)
=
2(1)
√
5 ± 57
.
=
2
The decimal approximations to these two solutions are p = −1.275 and p = 6.275.
18. We begin by using the distributive law. The resulting equation is quadratic in x, so we combine terms and set the expression
equal to zero. We have:
x(x + 5) − 2(4x + 1) = 2
x2 + 5x − 8x − 2 = 2
x2 − 3x − 4 = 0
(x − 4)(x + 1) = 0
x = 4 , x = −1.
19. We begin by using the distributive law. The resulting equation is linear in s, so we put terms with s on one side and terms
without s on the other. We have:
5(2s + 1) − 3(7s + 10) = 0
10s + 5 − 21s − 30 = 0
−11s − 25 = 0
−11s = 25
s = −2.273.
20. We begin by using the distributive law. The resulting equation is quadratic in t, so we combine terms and set the expression
equal to zero. We have:
2t(1.3t − 4.8) + 4.6(t2 − 8) = 5t(0.8t + 3.9) − 5(3t + 4)
2.6t2 − 9.6t + 4.6t2 − 36.8 = 4t2 + 19.5t − 15t − 20
3.2t2 − 14.1t − 16.8 = 0.
Since the quadratic expression does not easily factor, we use the quadratic formula. We have
√
−b ± b2 − 4ac
t=
2a
p
−(−14.1) ± (−14.1)2 − 4(3.2)(−16.8)
=
2(3.2)
√
14.1 ± 413.85
.
=
6.4
The decimal approximations to these two solutions are t = −0.976 and t = 5.382.
366
Chapter Nine /SOLUTIONS
21. This equation is quadratic in p. We use the quadratic formula. We have
√
−b ± b2 − 4ac
p=
2a
p
−5q ± (5q)2 − 4(1)(8qr 3 )
=
2(1)
=
−5q ±
p
25q 2 − 32qr 3
.
2
22. This equation is linear in q. We put terms with q on one side and terms without q on the other. We have
p2 + 5pq + 8qr 3 = 0
5pq + 8qr 3 = −p2
(5p + 8r 3 )q = −p2
q=
−p2
.
5p + 8r 3
23. This equation involves a power of r. We can solve it by isolating the r 3 and then raising both sides to the 1/3 power:
p2 + 5pq + 8qr 3 = 0
8qr 3 = −p2 − 5pq
−p2 − 5pq
r3 =
8q
r=
−p2 − 5pq
8q
1/3
.
24. This equation is quadratic in s so we start by combining terms so that one side is 0. We have
as − bs2 + ab = 2bs − a2
as − bs2 + ab − 2bs + a2 = 0
2
−bs + (a − 2b)s + (ab + a2 ) = 0.
To solve the quadratic equation ax2 + bx + c = 0, we use the quadratic formula:
√
−b ± b2 − 4ac
x=
2a
In the equation above, the coefficient of s2 is −b, the coefficient of s is (a − 2b), and the constant is (ab + a2 ). We
substitute this information and get
s=
=
−(a − 2b) ±
2b − a ±
p
(a − 2b)2 − 4(−b)(ab + a2 )
2(−b)
p
(a − 2b)2 + 4ab2 + 4a2 b
.
−2b
25. This equation is quadratic in a so we start by combining terms so that one side is 0. We have
as − bs2 + ab = 2bs − a2
2
a2 + as − bs2 + ab − 2bs = 0
a + (s + b)a + (−bs2 − 2bs) = 0.
SOLUTIONS to Review Problems for Chapter Nine
We use the quadratic formula:
367
√
b2 − 4ac
.
2a
2
In the equation above, the coefficient of a is 1, the coefficient of a is a + b, and the constant is (−bs2 − 2bs). We
substitute this information and get
−b ±
x=
a=
=
p
(s + b)2 − 4(1)(−bs2 − 2bs)
2(1)
−(s + b) ±
−s − b ±
p
(s + b)2 + 4bs2 + 8bs
.
2
26. This equation is linear in b so we start by putting terms with b on one side of the equation and terms without b on the
other. We have
as − bs2 + ab = 2bs − a2
−bs2 + ab − 2bs = −a2 − as
bs2 − ab + 2bs = a2 + as
(s2 − a + 2s)b = a2 + as
b=
a2 + as
.
− a + 2s
s2
27. We first use the distributive law. The resulting equation is quadratic in p, so we combine terms so that one side of the
equation is 0. We have
p(2q + p) = 3q(p + 4)
2pq + p2 = 3pq + 12q
2
p − pq − 12q = 0.
We use the quadratic formula:
√
b2 − 4ac
2a
p
−(−q) ± (−q))2 − 4(1)(−12q)
=
2(1)
p=
=
−b ±
q±
p
q 2 + 48q
.
2
28. We first use the distributive law. The resulting equation is linear in q, so we first put terms with q on one side and terms
without q on the other. We have
p(2q + p) = 3q(p + 4)
2pq + p2 = 3pq + 12q
p2 = pq + 12q
(p + 12)q = p2
q=
p2
.
p + 12
368
Chapter Nine /SOLUTIONS
29. We first use the distributive law. The resulting equation contains a power of H, so we start by isolating the power, then
raising both sides to the 1/5 power. We have
2[T ]V0 = 5V (H 5 − 3V0 )
2[T ]V0 = 5V H 5 − 15V V0
5V H 5 = 2[T ]V0 + 15V V0
2[T ]V0 + 15V V0
H5 =
5V
H=
2[T ]V0 + 15V V0
5V
1/5
.
30. We first use the distributive law. The resulting equation is linear in V0 , so we start by putting terms with a V0 on one side
of the equation and terms without a V0 on the other. We have
2[T ]V0 = 5V (H 5 − 3V0 )
2[T ]V0 = 5V H 5 − 15V V0
2[T ]V0 + 15V V0 = 5V H 5
(2[T ] + 15V )V0 = 5V H 5
V0 =
5V H 5
.
2[T ] + 15V