Unit 5 – Selected Answers
Intermediate Algebra (B)
SECTION 5.1A
1) “Find the zeros of the function”
means to find the x-value(s)
that make the function f(x) = 0.
(answers will vary)
2) The x-intercepts (answers will
vary)
3) # & type: 1 real root (rational)
zeros: x = 3
4) # & type: no real roots
zeros: none
5) # & type: 2 real roots (rational)
zeros: x = 0 or x = –2
6) solutions: x = 3 or x = –5
check: f(3) = (3)2 + 2(3) – 15 = 0
f(–5) = (–5)2 + 2(–5) – 15 = 0
both solutions create a zero
7) solutions: x = 0 or x = 4
check: f(0) = –2(0)2 + 8(0) = 0
f(4) = –2(4)2 + 8(4) = 0
both solutions create a zero
8)
9)
17)
solutions: x = –5 or x = –1
18)
solutions: x = –5 or x = 2
check:
f(–5) = (–5)2 + 3(–5) – 10 = 0
f(2) = (2)2 + 3(2) – 10 = 0
10) solutions: x ≈ –1.32 or x ≈ 8.32
11) solutions: x ≈ –1.67or x = –1.50
12) solutions: x ≈ –0.94 or x ≈ 2.34
13) solutions: x ≈ –3.12 or x ≈ 5.12
0.5(−3.12)
− (−3.12) − 8 ≈ 0
0.5(5.12)
− (5.12) − 8 ≈ 0
14) solutions: x ≈ –5.16 or x ≈ 1.16
(−5.16)
+ 4(−5.16) ≈ 6
(1.16)
+ 4(1.16) ≈ 6
15) solutions: x ≈ –1.16 or x ≈ 2.16
16)
solutions: x = –5 or x = –1
19)
2(−1.16)
− 2(−1.16) − 5 ≈ 0
2(2.16)
− 2(2.16) − 5 ≈ 0
solutions: x = –2 or x = 6
check:
f(–2) = –(–2)2 + 4(–2) + 12 = 0
f(6) = –(6)2 + 4(6) + 12 = 0
solutions: x = 0 or x = –8
solutions: x = –5 or x = –1
Answers
Intermediate Algebra (B)
~ page 1 ~
Unit 5 – Selected Answers
Intermediate Algebra (B)
20)
solutions: x = 0 or x = –8
21)
SECTION 5.1A (continued)
22) a) (16) f(x) = x2 + 6x + 5
x = –5, x = –1
(17) f(x) = 5x2 + 30x + 25
x = –5, x = –1
(18) f(x) = 3x2 + 18x + 15
x = –5, x = –1
b) (19) f(x) = x2 + 8x
x = 0, x = –8
(20) f(x) = (½)x2 + 4x
x = 0, x = –8
(21) f(x) = 2x2 + 16x
x = 0, x = –8
c) If one function is a multiple
of another function, then
they will have the same
solutions. (answers will
vary)
23) a)
b) 5 feet; the point (5, 5) is the
vertex of the graph of the
equation.
c) 10 feet; this is the x-intercept
of the graph of the equation
(that is greater than zero).
(answers will vary)
solutions: x = 0 or x = –8
1) a) Peter is correct. The
x-intercept gives the time
(t) when y = 0, which is
where the ball has a height
of 0 on the ground.
(explanations will vary)
b) Pablo’s ball hit the ground
first. It took 2.5 seconds
for it to hit the ground.
c) Peter’s ball hit the ground
second. It took 3.125
seconds for it to hit the
ground.
2) a) h(t) = –16t2 + 35t + 5.5
b) 24.5 feet
c) After 0.14 seconds and
then again after 2.05
seconds.
d) After 1.09 seconds.
Answers
Intermediate Algebra (B)
2)
3)
4)
5)
SECTION 5.1B
e) 24.64 feet is the maximum
that the football ever gets
into the air.
f) 2.33 seconds is when the
ball hits the ground
(the x-intercept).
Vertex: (1.56, 41.06). The
vertex reveals how long it
takes (1.56 seconds) for the
ball to reach the maximum
height (41.06 ft).
(explanations will vary)
t = 11 days, which is the
x-coordinate of the vertex.
a) 1.5 sections is the
maximum speed (the
x-coordinate of the vertex
of the graph)
b) 89 km/h (the y-coordinate
of the vertex of the graph)
5) c) 6.22 seconds (the
x-coordinate of the
x-intercept of the graph.
6) a) 50 is half the perimeter, so
one length plus one width
equals 50 feet of fencing.
(explanations will vary)
b) A(L) = L(50 – L)
A(L) = –L2 + 50L
c) (L, A(L)) = (25, 625)
when the length is 25, the
width is 50 – L = 25. So,
Area = L × W
25 ft × 25 ft
625 ft2
Area of the tomato patch is
625 square feet.
d) The tomato patch is a
square shape (25ft × 25ft).
~ page 2 ~
Intermediate Algebra (B)
7) a) According to #6d, Sharon
should make a square with
her 40 feet of fencing. So,
40 ÷ 4 = 10 feet on each
side.
b) A(L) = L(20 – L)
A(L) = –L2 + 20L
calc the max, shows a
vertex at (10, 100)
Unit 5 – Selected Answers
SECTION 5.1B (continued)
7 c) (L, A(L)) = (10, 100)
when the length is 10, the
width is 20 – L = 10. So,
Area = L × W
10 ft × 10 ft = 100 ft2
The area of the fenced play
area is 100 square feet.
SECTION 5.2A
16) (x + 5)(x – 2)
1) 2(x + 3)
2
2) 3(y – 3)
17) (x – 5)(x – 5) or (x – 5)
3) 7(a + 4)
18) (x – 4)(x + 3)
4) 12(3z – 1)
19) (3x + 4)(x – 1)
5) b(b + 1)
20) (2x – 3)(x + 4)
6) r(2 – r)
21) (2x – 3)( 2x – 3) or (2x – 3)2
7) t(9t + 1)
22) (6x – 1)(2x – 1)
8) n(4n – 5)
23) 2(x + 1)(x – 5)
9) 4h(h + 3)
24) 3(x + 1)(x + 6)
10) 9x(1 – 3x)
25) (a – 3)(a + 3)
11) 2a(a + 2)
26) (x – 4)(x + 4)
12) 4d(5d – 6)
27) (5x – 6)(5x + 6)
30) The x-intercepts have a
y-coordinate of zero.
Setting y = 0 and solving for
x gives the solution(s),
which are the x-intercept(s).
(answers will vary)
31) (4, 0) and (20, 0)
32) (1, 0) and (–10, 0)
33) a) an arrow should be drawn to
the location that the graph of
the height of the ball in
relationship to time touches
the x-axis.
33) b) y = 0 means the place where
the ball hits the ground
(where the height is zero).
28) y = (x + 1)(x + 2)
13) (x + 7)(x + 6)
14) (x + 3)(x + 3) or (x + 3)
8) Yes! The maximum height of
the orange is 19 feet (which is
the y-coordinate of the vertex
of the graph). Jim can grab
the orange either on the way
up for the orange or on its way
down (explanations will vary).
9) At approximately 15 miles per
hour will the car be able to
stop for a sign 30 feet away.
10) a) 141.56 feet
b) 5.79 seconds
2
29) y = (x + 7)(x – 7)
15) (x + 8)(x + 4)
1) x = 4 or x = –9
SECTION 5.2B
8) x = 1 or x = –9
2) x = 2 (double root)
9) x = –4 or x = –3
3)
x=−
3
5
or x =
4
2
10) x = 5 (double root)
12) x = –1 or x = 5
5) x = 2 or x = –2
13) x = 5 or x = 8
7) x = 0 or x = 4
Answers
Intermediate Algebra (B)
5
1
or x =
4
2
16) x = −
1
or x = –2
5
17) x = −
1
or x = 8
2
11) x = –1 or x = 4
4) x = 3 or x = –3
6) x = 0 or x = 3
15) x = −
14) x = −
4
or x = –2
3
~ page 3 ~
Intermediate Algebra (B)
Unit 5 – Selected Answers
1) The legs of the triangle have
lengths of 3 units and 4 units.
2) The legs of the triangle have
lengths of 5 units and 12 units.
3) The value of x is 10 (side
lengths are 9 units and 12
units).
SECTION 5.2C
4) The value of x is 11 (side
lengths are 10 units and 12
units).
5) The two numbers that satisfy
the situation are 8 and 15.
6) The dimensions are 14 feet by
48 feet.
1) a) 2 5
4) a) 5
7) The two positive numbers are
5 and 3.
8) The two negative integers are
–4 and –6.
9) (x = 2) You can frame a
2ft × 2ft square picture.
SECTION 5.2D
8) a) 2a + 5
b) 4 3
b) 3 10
b) 2 + 5 2
c) 10 2
c) 3 2
c) −7 − 5 6
d) 6 5
d) 12 15
d) −5 + 6 10
e) 10 6
e) –80
f) 42 2
f) −6 42
2) a) 9
b) 6 + 11 2
5) a) 15 + 5 7
c) 2 − 10 7
b) −6 + 3 3
d) 5 − 5 2
c) 2 + 8 3
e) 3 − 5
d) 2 5 + 3 10
f) 11 + 8 3
3) a)
e) −6 − 3 2
f) −5 3 + 2 6
6) a) 19 + 8 3
b) 22 − 12 2
c) 47 − 26 3
9)
 3 + 57
2 
4

2
?

 3 + 57 
 − 3 
 − 6 = 0
4



?
 9 + 6 57 + 57   3 + 57 
2 
 − 3 
 − 6 = 0
16
4

 

?
 66 + 6 57  −9 − 3 57
2 
−6=0
+

16
4


?
 33 + 3 57  −9 − 3 57
2 
−6=0
 +
8
4


?
33 + 3 57 −9 − 3 57
+
−6=0
4
4
?
33 − 9 + 3 57 − 3 57
−6=0
4
?
24
−6=0
4
?
6−6=0
d) 76 + 24 2
0=0
e) 132 − 48 7
3) b)
f) 42 − 7 11
7) a) yes, this x-value is a
solution
b) no (x = 6 + 2 ) is not a
solution to the equation.
c) yes, (x = 5 − 3 5 ) is a
solution to the equation.
Answers
Intermediate Algebra (B)
~ page 4 ~
Intermediate Algebra (B)
1)
2)
3)
4)
5)
6)
x = –2
x = –4
x = –5
x = –5
x = –7
x = –1
or
or
or
or
or
or
x = 2 ( x = ± 2)
x = 4 ( x = ± 4)
x = 5 (x = ±5)
x=7
x=1
x=5
1) The error occurs when Omar
tried to add 3 to both sides
before he takes the square root
of both sides. The correct
solutions are x = 6 or x = 0
2) Yes, both of the values given
are solutions.
3) Yes, both of the values given
are solutions.
4) Yes, both of the values given
are solutions.
5) No, neither value given is a
solution.
Unit 5 – Selected Answers
SECTION 5.2E
7) It takes 2 seconds to reach the
ground.
8) A sign should be posted
stating the maximum speed is
28 km per hour.
SECTION 5.2F
6) Both answers are correct. The
balloon could have hit a 6 foot
tall student on the way up
after 0.15 seconds, or it could
have hit the student on its
descending path after 1.15
seconds.
9) The other leg of the triangle
Andy created is 9 feet long.
7)
x = ±3 2
8) x = 1 ± 4 2
9) x = ±2
10) x = –2 or x = 4
11) x = ±5 3
12) x = –2
13) x = –2 or x = 4
14) x = ±2 2
15) t ≈ 2.41 seconds
16) x ≈ 49.74 miles
(explanations will vary)
1) a) − 1
b) 1
c) i
d) 1
e) i
SECTION 5.2G
2) c) −2 + 2i
d) 11 + 3i
e) 9 − 3i
f) 12 − 19i
3) a) −6
f) −i
b) 12
g) 5i
c) 24i 2
h) 12i
d) −60i
i) 10i 7
e) 48i
j) 14i 2
f) 18 2
k) − 4i 10
l) 50i
2) a) 9 + 4i
b) 13 − 6i
Answers
Intermediate Algebra (B)
4) (all answers in #4 may vary)
a) (1) in method 1, Kasem
never used the definition on
i = −1 to rewrite either
radical. Method 3 uses the
definition correctly.
(2) method 1 used the
product property of radicals
incorrectly; only if a and b
are both nonnegative with an
even index, is
a ⋅ b = a ⋅ b . Method 2
uses the product property of
radicals correctly.
b) (1) both methods combine
4 ⋅ 9 = 36 instead of
simplifying each separately.
(2) both methods used the
fact that
36 = 6
~ page 5 ~
Intermediate Algebra (B)
4) c) (1) from line 3 to line 4,
multiplication was done in a
different order by
commutative property for
multiplication.
(2) method 3 simplified the
radicals
4 and 9 ;
method 2 multiplied them
together.
d) (1) both used the definition
of i = −1 whenever there
−1 .
(2) both used the fact that
was a
2
i = −1
5) a) −2 3
b) −12 10
c) 63 2
Unit 5 – Selected Answers
SECTION 5.2G (continued)
6) a) 5 − 15i
9) c) Yes
2
c) 6 − 60i
(1 − 3i 5 ) − 2 (1 − 3i 5 ) + 48 = 2
(1 − 3i 5 )(1 − 3i 5 ) − 2 + 6i 5 + 48 = 2
d) 12 − 6i
1 − 6i 5 + ( 9 ) ( −1) ( 5 ) − 2 + 6i 5 + 48 = 2
e) 10 + 40i
10) a) 3 + 4i
b) 9 + 3i
2=2
f) 2 5 + 2i 10
b) 3 + 5i 2
c) −5 − 7i 6
7) a) −16 + 30i
d) −6i 3
b) 36 − 48i
11)
c) 7 − 30i 2
2
?
 3+i 
 3+i 
2
 −6
+8=3
 2 
 2 
8) a) −68 + 24i 2
b) −36 − 48i 7
c) −24 − 7i 11
9) a) Yes
−2(5)
+ 6 = 56
56 = 56
b) Yes
(4 + 3)
− 8(4 + 3) + 30 = 5
5=5
?
 9 + 6i + i 2 
2
 − 3(3 + i ) + 8 = 3
4


?
 9 + 6i + ( −1) 
2
 − 9 − 3i + 8 = 3
4


?
8
6i 
+

2
 − 1 − 3i = 3
 4 
?
 4 + 3i 
2
 − 1 − 3i = 3
 2 
?
4 + 3i − 1 − 3i = 3
3=3
SECTION 5.2H
(#23-26, verification of solutions should
be shown by student.)
1) 3i
2) 35i 2
3) 2i 3
4) 30i
5) 33i 2
6) 11i
7) –1
8) –9
9) –25
10) –1
11) −16i − 30i
12) −14 + 48i
13) −40 + 28i
14) Yes
−2(3)
+ 3 = 21
21 = 21
Answers
Intermediate Algebra (B)
15) Yes
(5 + 4) − 5 − 1 = −17
−17 = −17
16) Yes
(2)
+ 11 = 7
7=7
17) Yes
(−2 + 5) + 2 = −25
−25 = −25
18) = ±4√3
19) = 1 ± 2√6
20) = ±3
21) = −1 ± 3
22) ≈ 1.86 seconds. The balloon
lands on the ground after
approximately 1.86 seconds.
(–0.96 does not work when
verified)
23) = ±5√5
24) = −7
25) = 2 ± 3
26) = ±√6
27) s ≈ 7.071 inches;
The length of one side of the
square is approximately 7.071
inches.
28) = ±2. With these
solutions being imaginary, it
reinforces the reason why the
graph does not cross the
x-axis. There are no real
solutions, there are two
complex solutions.
~ page 6 ~
Unit 5 – Selected Answers
Intermediate Algebra (B)
1)
x = 7 or x = −3
7)
SECTION 5.2I
x = 1 or x = −29
2)
x = 0 or x = −8
8)
x = 14 or x = −4
3)
x = 18 or x = −4
9)
x=6
4) From line 1 to line 2, the
person forgot to add 25 to
both sides of the equation.
5) x = 4 or x = −10
6)
10) = 27 or = 3
11) = 22 The two numbers that
satisfy the conditions are 22
and 24.
12) = 5 The length of each side
of the original garden is 5
yards and the area of the
garden is 25 square yards.
13) = 2 The ball was in the air
for 2 seconds when it reached
a height of 64 feet in the air.
x = 22 or x = −4
SECTION 5.2J
1)
x = −7 ± 15
2)
x =9±2 3
3)
x = 5± 4 2
−5 ± 53
4)
x=
5)
x=
6)
x = −14 ± 6 3
7)
x=
2
7 ± 37
2
7 ± 73
4
1)
x = 5 ± 12i
2)
x = 7 ± 2i 6
3)
4)
x = −13 ± 8i 2
5)
x = 12 ± 6i 2
x = −6 ± 2i
−1 ± i 15
2
7) x = 2 ± 4i
6)
8)
x=
x=
3 ± i 21
10
−3 ± i 31
8
10) x ≈ 250.81 ; Emma hits the
golf ball approximately
250.81 yards.
9)
8)
x=
9)
x=
−1 ± 22
2
1± 2
12) t = 6 ; The arrow will strike
the ground after 6 seconds.
13) a) h ( 2 ) = 424 ; The bottle
rocket is 424 feet high after
2 seconds.
b) t ≈ 13.13 or t ≈ 1.87 ; The
rocket will be at 400 feet in
the air at two different
times. Once on the way
‘up’ and once on the way
‘down’ during its flight.
3
10) When y > 0, there will be 2
real solutions.
When y = 0 there will be 1
real solution.
11) x = 1.5 ; The width of the
pool is 15 meters and the
length is 19 meters.
SECTION 5.2K
1
1
11) x = or x = 1 ; At exactly
2
2
½ second and 1½ seconds, the
tape measure is at exactly 17
feet above the ground. Gail
can catch the tape measure
anywhere between ½ second
and 1½ seconds after
Veronica tosses the tool.
13) Creates the quadratic formula:
ax 2 + bx + c = 0
ax 2 + bx + c
0
=
a
a
bx c
2
x + + =0
a a
bx
c
x2 +
=−
a
a
2
x2 +
bx  b 
c b2
+
 =− + 2
a  2a 
a 4a
2
b 
b 2 4ac

x+
 = 2− 2
2a 
4a
4a

12) When y < 0 there will be 2
imaginary solutions for a
quadratic equation.
2
b 
b 2 − 4ac

x+
 =
2a 
4a 2

x+
x=
x+
b
2a
x=−
=−
b
2a
−
b
2a
b 2 − 4ac
=
or x +
2a
b 2 − 4ac
b 2 − 4ac
2a
b
2a
or x = −
2a
x=−
b
2a
±
=+
b
2a
+
b 2 − 4ac
2a
2
x=
Answers
Intermediate Algebra (B)
−b ± b − 4ac
2a
~ page 7 ~
b 2 − 4ac
2a
b 2 − 4ac
2a
Intermediate Algebra (B)
Unit 5 – Selected Answers
SECTION 5.2L
1) If the quadratic equation does
not factor, the roots are either
irrational or imaginary.
(Answers will vary)
2) a: 2
b: –5
c: –3
1
2
solve by factoring? Yes,
b 2 − 4 ac (49) is a perfect
square number.
x = 3 or x = −
3) a: 1
b: –7
c: 9
7 ± 13
x=
2
solve by factoring? No,
b 2 − 4 ac (13) is not a perfect
square number.
1) 2i 3
2) 2 6
3) 4i
4) x = 1 or x = −3
3
5) x = −
2
3 ± 2i
6) x =
2
−6 ± i 11
4
8) x = −1 ± i 5
9) x = 3 or x = −5
4) a: 5
x=
b: 3
−3 ± 29
c: –1
x=
10
solve by factoring? No,
b 2 − 4 ac (29) is not a perfect
square number.
5) a: 1
b: 1
c: –1
−1 ± √5
2
solve by factoring? No,
b 2 − 4 ac (5) is not a perfect
square number
=
SECTION 5.2M
10) Error on line 6, need to divide
both terms of the numerator
by the denominator value of 2;
x = −3 ± i
11) x = −3 ± 7
12) x = 2
13 ± i 11
10
14) a) 2.6 seconds
13) x =
7) x =
1) » Use a graphing utility to find
real zeros.
» Factor and use the zero
product property.
» Square root property
» Complete the square
» Quadratic formula
Answers
Intermediate Algebra (B)
6) a: 9
SECTION 5.2N
2) Graph: The solutions are the
x-intercepts of x = 3, x = –2
7) a: 2
x=
b: 6
−1 ± 2
c: –1
3
b: 3
−3 ± 17
c: –1
4
8) 1.68 seconds
9) a) 22 seconds
b) 11 seconds
14) b) Since t = 0 means time
when object is initially
thrown, positive time
means time after the
throw was started.
Negative time implies
time before the object was
thrown, which doesn’t
make sense here.
(Answers will vary)
c) The object was at a height
of 25 feet after 0.75
seconds and again at 1.75
seconds.
Factor:
( x − 3 )( x + 2 ) = 0
x = 3 or x = −2
Quadratic Formula:
1 ± 25
2
1± 5
x=
2
x = 3 or x = −2
solutions:
x = 3 or x = −2
x=
~ page 8 ~
Intermediate Algebra (B)
2) method:: The factoring method
is the fastest and yields rational
solutions with fewer possibilities
of careless errors.
(explanations may vary)
3) Graph: The solutions are the xintercepts of x ≈ ±1.73
Square Root:
− 3 = 0
= √3
|| = ±√3
Quadratic Formula:
0 ± 12
x=
2
±2 3
x=
2
x=± 3
solutions: x = ± 3
method: The square root
method. There is no linear term
(the x term). The b-value is equal
to zero. (Answers will vary)
4) a) Graphing calculator (but it
can’t be used to find
complex/imaginary solutions)
b) Completing the Square and
Quadratic Formula
c) Factoring or Square Root
property
5) Factoring – it’s a trinomial that
factors nicely. (method and
explanation may vary)
1
x=
or x = −3
2
Answers
Intermediate Algebra (B)
Unit 5 – Selected Answers
SECTION 5.2N (continued)
6) Factoring – it is a binomial with a
common factor. (method and
explanation may vary)
x = 0 or x = 1000
7) Factoring – it is a trinomial with a
leading coefficient of 1. (method
and explanation may vary)
x = −9 or x = 2
8) There are two scenarios for this
situation:
(1) Numbers 8 and 9
(2) Numbers –8 and –9
9) width: 7 in. length: 10 in.
10) Suzie needs 80 feet of fencing.
11) Mike needs 40 feet of fencing.
12) Solutions: x = ±3
Methods included here may vary
Method 1: Factoring
2 ( x − 3 )( x + 3 ) = 0
Method 2: Square Root Property
2 x 2 + 3 = 21
2 x 2 = 18
x2 = 9
continued …
Why? Using square roots has a
less chance for careless sign
error. Also, we have been using
square roots to solve longer than
using factoring to solve.
(methods may vary)
13) Solutions: x = 25 ; so the
width is 25 m and length is 35 m
Method 1:Complete the Square
x ( x + 10 ) = 875
x 2 + 10 x + ____ = 875 + ____
13) Why? Complete the Square was
whole number, and the square
root property is well engrained.
Factoring might have taken
longer to find 2 integers with a
product of –875.
(answers/methods may vary)
14) x =
−3 ± i 21
3
be verified
15) x =
6 ± 39
3
be verified
16) x = −3
; solutions should
; solutions should
(−3)
+ 6(−3) + 9 = 0
0=0
17) x = ±
i
9
81 # $ + 1 = 0
9
0 = 0 and
− 81 # $ + 1 = 0
9
0=0
18) a) The ball was in the air 5 sec.
b) The ball reaches the max
height at 2.5 seconds.
c) The max height is 100 feet.
d) Graph a second line at y2 = 60
and use “calculate intersect”
to get the x-coordinate that
creates a function value of 60.
The ball is at 60 feet again at
approximately 4.08 seconds.
x 2 + 10 x + 25 = 900
( x + 5)
2
more efficient since ! was a
= 900
continued …
Note that –35 is extraneous.
Method 2: Factoring
x ( x + 10 ) = 875
x 2 + 10 x − 875 = 0
( x − 25 )( x + 35 ) = 0
continued …
~ page 9 ~
Intermediate Algebra (B)
Unit 5 – Selected Answers
SECTION 5.2O
(1) – (5) (answers will vary)
Square Root Property:
Eq’n #4:
( x − 3)
2
=8
6) a) x = 7 or x = 1
b) graph:
Graphing:
Eq’n #5:
h ( t ) = −16t 2 + 24t + 4755
Solution(s): x = 3 ± 2 2
3 + 2√2 − 3! = 8
8=8
and
3 − 2√2 − 3! = 8
8=8
Factoring:
Eq’n #2: x 2 − 2 x − 15 = 0
Solution(s): x = 5 or x = −3
(5)
− 2(5) − 15 = 0
0=0
and
(−3)
− 2(−3) − 15 = 0
0=0
Factoring:
Eq’n #3: x 2 + 12 x = 20
Solution(s): x = −6 ± 2 14
Solution(s): t ≈ 18.005
It took approximately 18
seconds for the rock to hit the
ground.
−16(18.005)
+ 24(18.005) + 4755 = 0
0=0
−6 + 2√14 + 12−6 + 2√14 = 20
20 = 20
and
−6 − 2√14 + 12−6 − 2√14 = 20
20 = 20
Quadratic Formula:
Eq’n #1: 2 x 2 + 7 x − 15 = 0
3
Solution(s): x = or x = −5
2
3
3
2 # $ + 7 # $ − 15 = 0
2
2
0=0
And
2(−5)
+ 7(−5) − 15 = 0
0=0
1) In the quadratic formula,
−b ± b 2 − 4ac
, the
2a
discriminant is the value of the
expression b 2 − 4ac that is
under the radical. This number
is used to determine the
number and type of solutions
of a quadratic equation.
x=
Answers
Intermediate Algebra (B)
SECTION 5.3A
2) discriminant: –4
number of solutions: 2
type: imaginary
3) discriminant: 16
number of solutions: 2
type: real, rational
4) discriminant: 0
number of solutions: 1
type: real, rational
c) The intersection points of
the 2 graphs in part b) are
the solutions for x in part a).
9 = (7 − 4)
9=9
and
9 = (1 − 4)
9=9
7) a) The 4.5 feet is how far the
hose’s nozzle is above the
ground where the water
begins to shoot out.
b) 38.55 feet
c) No. At 27.89 feet from the
nozzle, the stream would hit
the top of the 6 foot fence.
However, at the 28 foot
distance, the water height is
lower, reaching only 5.96
feet high.
8) (answers will vary)
Xmin: –30
Xmax: 30
Ymin: –400
Ymax: 1000
5) discriminant: 217
number of solutions: 2
type: real, irrational
6) discriminant: 0
number of solutions: 1
type: real, rational
7) discriminant: –52
number of solutions: 2
type: imaginary
~ page 10 ~
Intermediate Algebra (B)
8) discriminant: 1
number of solutions: 2
type: real, rational
9) discriminant: –7
number of solutions: 2
type: imaginary
10) the second line of ‘work’, the
negative 6 must be in
Unit 5 – Selected Answers
SECTION 5.3A (continued)
11) a) 40000 = 0.003 x 2 + 12 x + 27760
b) 0.003 x 2 + 12 x − 12240 = 0
c) 290.88
d) Yes, the discriminant > 0,
which yields two real solutions.
Only the positive solution
would pertain to this story.
12) d = 977. Yes, the discriminant
is greater than zero and yields
two real solutions.
13) d = –280,000. No, since
discriminant < 0 there are no
real solutions.
2
parentheses ( −6 ) − 4 (1) ( 5 )
discriminant: 16
# of solutions 2
type: real, rational
1) a)
b)
c)
d)
36
2
real, rational
= −2 or = 4
e)
f) (1, –9)
g) minimum
h) (0, –8)
i) all real numbers
j) ( ≥ −9
1) a)
b)
2) a)
b)
3) a)
b)
Not a solution
Solution
Solution
Not a solution
Not a solution
Not a solution
Answers
Intermediate Algebra (B)
2) a)
b)
c)
d)
SECTION 5.3B
36
2
real, rational
= ±3
e)
f) (0, 9)
g) maximum
h) (0, 9)
i) all reals
j) ( ≤ 9
SECTION 5.4A
4) (test points may vary)
Use (0, 0)
Is 0 ≤ −4? No, so points
outside the parabola are
solutions.
3) D
4) discriminant = 28
2 solutions, real, irrational
5) discriminant = 0
1 solution, real, rational
6) discriminant = –11
2 imaginary solutions
7) a) negative discriminant
b) zero discriminant
c) positive discriminant
8) Discriminant is a positive
perfect square number
2 real, rational solutions
Possible equation:
y = − ( x − 3 )( x − 7 )
y = − x 2 + 10 x − 21
(Answers may vary)
5)
(test points may vary)
Use (0, 0)
Is 0 ≥ −3? Yes, so points
outside the parabola are
solutions.
~ page 11 ~
Unit 5 – Selected Answers
Intermediate Algebra (B)
SECTION 5.4A
17)
6) (test points may vary)
Use (0, 0)
Is 0 > 6? No, so graph
outside the parabola.
(continued)
20)
y
8
6
4
2
-6
-4
-2
2
x
-2
21)
y
4
7)
8)
9)
10)
11)
12)
13)
14)
15)
16)
2
<
≤
>
C
A
F
E
B
D
18)
y
-10 -8
-4
-2
2 x
-2
4
-4
2
-2
-6
2
4
6
x
22)
y
-2
y
4
-4
4
2
2
-4
-6
-2
19)
2
4
-2
y
-2
4
-4
-2
x
2
x
23)
y
8
2
-6
6
-4
6
-4
4
-2
8
x
2
2
6
x
-2
4
2
-6
-4
-2
-2
Answers
Intermediate Algebra (B)
~ page 12 ~
Unit 5 – Selected Answers
Intermediate Algebra (B)
1) a) When < −4or > 2
b) −4 < < 2
c) graph
SECTION 5.4B
5) < −5or > 5
9) x 2 − 8 x − 20 > 0 x < −2 or x > 10
y
5
Function
values are
MORE than 0
above x-axis
Function
values are
MORE than 0
above x-axis
-5
5
10
-5
2) a) −3 < < 2
b) < −3or > 2
c) graph
6) ≤ 0or ≥ 4
10) x 2 − 12 x + 27 < 0
3 < < 9
Function
values are
MORE than 0
above x-axis
Function
values are
LESS than 0
below x-axis
7) 2 < < 5
-5
Function
values are
LESS than 0
below x-axis
8)
4) Answers will vary, but
x-intercepts and concavity must
match. Example:
Intermediate Algebra (B)
8 x
Function
values are
MORE than 0
above x-axis
-5
5 − 73
5 + 73
≤x≤
2
2
Function
values are
LESS than 0
below x-axis
Answers
0
11) ≤ or ≥ 5
y
5
3) Answers will vary, but
x-intercepts and concavity must
match. Example:
12)
2
< x<5
3
Function
values are
LESS than 0
below x-axis
~ page 13 ~
x
Intermediate Algebra (B)
13) $20 < < $100 (line is
dashed)
x-intercepts are 20, 100 and
vertex is at (60, 1600)
14) 1 ≤ ≤ 3 seconds
x-intercepts are at 1 and 3,
vertex is at (2, 16)
Unit 5 – Selected Answers
SECTION 5.4 B (continued)
15) a) 0 ≤ ≤ 8.03ftand
≥ 18.45ft. The
x-intercepts are at 8.03
and 18.45 and the vertex
is at (13.24, 1.47)
b) No. From 15 feet away,
the height of the ball
would be 9.3 feet, so the
ball would go over the top
of the net. Also, from the
graph and algebraic
solution inequalities
above, 15 is not in the
correct ranges
(explanations may vary).
16) 4 ≥ 2 inches. Values in the
range 4 ≤ −2 are extraneous.
x-intercepts are at 2 and –2,
the vertex is at (0, -5920)
17) Driver’s age is between 56.6
years and 70 years (inclusive
on 70).
Unit 5 – Review Material
1) Answers will vary
a) There is no x-term, such as
= 4orif it is in vertex
2
form, like 3 ( x − 2 ) + 5 = 0
b) If it is unfactorable and
a = 1 and b is even
c) a, b, c are all integers and
are fairly small numbers
d) If it is unfactorable and
a, b, and c are larger
numbers and/or decimals.
2) Answers may vary
Suggest A, J and K for square
root method
Suggest B, F and I for
factoring
Suggest C, D, and E for
completing the square
Suggest G, H and L for
quadratic formula
Answers
Intermediate Algebra (B)
= ±3√2
= 1 ± 2√3
= 13or = −1
= −10 ± 2
= 3 ± 2√6
= 2or = −6
2
5)a) = − or = −5
3
3
b) = − or = 5
2
2
c) = −
3
−3 ± √3
6)a) =
3
4 ± √2
b) =
6
3
c) = 2or =
5
3) a)
b)
c)
4) a)
b)
c)
7)a) = −4 ± 5√2
1
b) = 1or = − 2
c) = −2 ± √5
7 ± 2√3
d) =
14
8) t = 5 seconds
9) 5.6 seconds
10) a) Disc. is negative. There
are two imaginary
solutions.
b) Disc. is zero. There is one
real, rational solution.
c) Disc. is positive. There are
two real solutions.
~ page 14 ~
Intermediate Algebra (B)
11) a) Disc. is –16. There are
two imaginary solutions.
b) Disc. is 261. There are
two irrational solutions.
c) Disc. is 0. There is one
real, rational solution.
12) B
13) −5 < < −1
Answers
Intermediate Algebra (B)
Unit 5 – Selected Answers
Unit 5 – Review Material (continued)
14)
• Algebraically find the
x-intercepts.
• Sketch the graph of a
parabola that has these
x-intercepts and opens up if
a > 0 or down if a < 0. Also
determine dashed or solid
line.
• Identify the x-values for
which the graph lies below
the x-axis (#15a) or above
(or on) the x-axis (#15b).
• For ≤ or ≥ include the
x-intercepts in the solution.
15) a) −2 < < 1 Function
values less than 0 (below
x-axis)
b) ≤ −8or ≥ −4
Function values are ≥ 0 (on
or above the x-axis)
16) Graph B
17) ≤ −1or ≥ 4
~ page 15 ~
Intermediate Algebra (B)
1. a) x < –1.69, x > 2.36
b) –1.69 < x < 2.36
c) (–1.69, 12.6) Relative Max
( 2.36, –20.75) Relative Min
These occur at turning points
d) (–3, 0), (0, 0), (4, 0)
e) (0, 0)
f) none
2. a) i) 1 ii) –1 iii) –1 iv) 1
b) If a is positive, the ends
have a positive slope If a is negative, the ends
have a negative slope 
c) They all either look like
or
. Some have
more of a squiggle in the
middle, with relative
maximums and minimums.
They differ in the number of
times each graph crosses the
x-axis (answers may vary).
3. a) They are all using different
windows, so they are seeing
different portions of the
graph. (answers may vary)
b) Meng’s is the best because
you can see all the key
features: max, min,
intercepts, end behavior.
(answers may vary)
4. Sign: Positive
Domain:ℝ
Range: ℝ
Rel Min: approx. (0.5, –8.5)
Rel Max: (–5.5, 2)
Increasing: x < –5.5 and x > 0.5
Decreasing: –5.5 < x < 0.5
x-int(s): (–7, 0), (–4, 0), (3, 0)
y-int: approx. (0, –8.33)
5. Sign: Negative
Domain:ℝ
Range: ℝ
Rel Min: approx. (–4.1, –1.1)
Rel Max: (0, 6)
Increasing: –4.1 < x < 0
Decreasing: x < –4.1 and x > 0
x-int(s): (2, 0), (–3, 0), (–5, 0)
y-int: approx. (0, 6)
Answers
Intermediate Algebra (B)
Unit 6 – Selected Answers
SECTION 6.1A
6. Sign: Positive
Domain:ℝ
Range: ℝ
Rel Min: approx. none
Rel Max: none
Increasing: ℝ
Decreasing: none
x-int(s): (0, 0)
y-int: approx. (0, 0)
7. Table
#
#1
#3
#5
x= (–5,0)
(0,0)
(3,0)
(1,0)
(3,0)
(7,0)
y= (0,-5.5)
(0,0)
(0,1)
Max (5.2, 2)
(1,4)
None
Min (–1.8, –7) (3,0)
None
Inc
x<1
–1.8<x<5.2
None
x>3
Dec x < –1.8
Dec. over
1<x<3
x > 5.2
entire dom.
Continued…
#4
#6
#2
(–8, 0)
(0,0)
(–2.4, 0)
(–2, 0)
(9, 0)
(0,0)
(0, 4)
(0, 3.7)
None
(–1.2, 7)
(4.5, 9.5)
None
(1.2, 0.9) (–5.5, –3.2)
Inc. over
x > 1.2
–5.5<x<4.5
entire dom. x < –1.2
x < –5.5
None
–1.2<x<1.2
x > 4.5
8. Explanations will vary.
a) Yes – if before year 1993
the number of acres was less
than 74,630 and declining
each previous year, then the
end behavior would be (↙↗)
b. Yes – if sometime after
1996 the number of bales
was > 9400 thousand and
increasing each succeeding
year, the end behavior
would then be (↙↗).
~ page 16 ~
Intermediate Algebra (B)
Unit 6 – Selected Answers
SECTION 6.1A
(continued)
9. Answers will vary but will all
be of the form y = x3.
10. Answers will vary but will all
be of the form y = –x3.
8. c) Yes – if before the year
1993 the yield per acre
was < 5.22 thousand and
decreasing each previous year,
the end behavior would then
be (↙↗).
1. Sign: Positive
End: ( )
Domain: ℝ
Range:ℝ
Rel Min: (1.29, –1.30)
Rel Max: (–1.29, 7.30)
Inc: x < –1.29 and x > 1.29
Dec: –1.29 < x < 1.29
x-int(s):(–2.49, 0),(0.66, 0),(1.83, 0)
y-int(s): (0, 3)
2. Sign: –Negative
End: ( )
Domain: ℝ
Range:ℝ
Rel Min: none
Rel Max: none
Inc: none
Dec: ℝ
x-int(s): (0.538, 0)
y-int(s): (0, 3)
3. Explanations may vary.
a determines the end behavior. If
a > 0 the end behavior is ( )
and if a < 0 the end behavior is
( ). d is the y-intercept.
4. Sign: Positive
End: ( )
Domain: ℝ
Range:ℝ
Rel Min: (2.08, –1.21)
Rel Max: (–2.08, 6.01)
Inc: x < –2.08 and x > 2.08
Dec: –2.08 < x < 2.08
x-int(s): (–4, 0), (1, 0), (3, 0)
y-int(s): (0, 2.4)
Answers
Intermediate Algebra (B)
SECTION 6.1B
5. Sign: –negative
End: ( )
Domain: ℝ
Range:ℝ
Rel Min: (–4.46, –8.31)
Rel Max: (2.46, 8.31)
Inc: –4.46 < x < 2.46
Dec: x < –4.46 and x > 2.46
x-int(s): (5, 0), (–7, 0), (–1, 0)
y-int(s): (0, 3.5)
6. Explanations may vary.
The x-intercepts occur at m, n,
and p. The y-intercept is
:(−;)(−)(−<). a
determines the end behavior.
If a > 0 the end behavior is
( ), and if a < 0 the end
behavior is ( ).
7. Sign: Positive
End: ( )
Domain: ℝ
Range:ℝ
Rel Min: none
Rel Max: none
Inc: ℝ
Dec: none
x-int(s): (–1.41, 0)
y-int(s): (0, 23)
8. Sign: Negative
End: ( )
Domain: ℝ
Range:ℝ
Rel Min: none
Rel Max: none
Inc: none
Dec: ℝ
x-int(s): (3.41, 0)
y-int(s): (0, 60.5)
9. Explanations may vary.
k moves the graph up and down;
h moves the graph left and right.
a stretches the graph. If a is
negative, the graph always
slopes down. If a is positive
the graph always slopes up.
There will be no maximum or
minimum values.
10) Sign: Positive
End: ( )
Domain: ℝ
Range: ( ≥ −4.3
Rel. Min. (–1.9, –4.3), (1.1, 1.95)
Rel. Max. (0, 4)
Inc: −1.9 < < 0, > 1.1
Dec: < −1.9, 0 < < 1.1
x-int: (–2.4, 0), (–1, 0)
y-int: (0, 4)
~ page 17 ~
Intermediate Algebra (B)
11) Sign: Positive
End: (↙↗)
Domain: ℝ
Range: ℝ
Rel. Min. (–0.23, –2.03), (4.5, –2.6)
Rel. Max. (–4.1, 7.3), (2.2, –0.7)
Inc: < −4.1,
−0.23 < < 2.2, > 4.5
Dec: −4.1 < < −0.23,
2.2 < < 4.5
x-int: (–5.3, 0), (–1.7, 0), (5.5, 0)
y-int: (0, –2)
12) Sign: Positive
End: (↖↗)
Domain: ℝ
Range: ( ≥ −5
Rel. Min: (3, –5)
Rel. Max: None
Inc: x > 3
Dec: x < 3
x-int: (0.8, 0), (5.2, 0)
y-int: (0, 11.2)
13) Sign: Positive
End: (↖↗)
Domain: ℝ
Range: ( ≥ −8.96
Rel. Min: (–2.45, –6.21),
(1.29, –8.96)
Rel. Max: (–1, 0)
Inc: –2.45 < x < –1, x > 1.29
Dec: x < –2.45, –1 < x < 1.29
x-int: (–3, 0), (–1, 0), (4, 0)
y-int: (0, –3.84)
14) Explanations may vary.
Even degree: end behavior is
either (↖↗) or (↙↘).
Odd degree: end behavior is
either (↙↗) or (↖↘).
15) a) 4272.9 million ft2 when
t = 18, since 1990 is 18
years after 1972.
b) Increasing. Explanations
may vary, but graph is
always increasing on this
interval.
c) No. When 0 < t < 24, S
increases each year.
Answers
Intermediate Algebra (B)
Unit 6 – Selected Answers
SECTION 6.1B (continued)
15) d) Domain: 0 ≤ ≤ 24
where t is the number of
years since 1972.
Range: 1700 ≤ @ ≤ 5050.5
S is the retail space (in
millions of ft2) over that
time period.
16) a) Domain: 0 ≤ ≤ 23,
Number of years from 1980
to 2003.
Range: $7.70 ≤ A ≤ $30.99,
monthly rate for cable TV in
that time period.
b) In the year 2003, the
maximum rate was $31.06
(23.8, 31.06). After that, the
rates decrease over time…
Not the case! For 2014 this
model would yield a rate of
$14 per month!
c) 0 ≤ ≤ 23, cable TV rates
were always increasing
from 1980 to 2003.
d) When t > 23.8, indicating
the rates started declining in
the year 2003.
e) A(3) = $8.55
17) Explanations may vary
a) L(18) = 50.7 in.
H(18) = 52.3 in.
50.7” < normal height < 52.3”
b) No. After a certain age, a
heifer does not get any
taller.
c) 6.6 mo. < age < 8.3 mo.
Draw a horizontal line
y = 43 and use “Calc.
Intersect” on both curves
L(8.3) = 43 and H(6.6) = 43.
d) At the point of inflection,
the increasing height starts
to level off before the graph
starts to increase again.
This occurs when the height
is near 55", so probable age
would be
30 mo. < age < 31.6 mo.
18) Explanations may vary
a) Domain: 0 < x < 6
Width > 0 and 12 – 2x > 0 ⟹ < 6
Range: 0 < V < 194.07
Volume > 0, and within
given domain the max.
volume is 194.07
b) 194 in3 within the
acceptable domain
0 < x < 6. The max volume
occurs at (2.26, 194.07)
c) Approximately 2.25 in2
19) Explanations may vary.
a) y-int. = 10. The roller
coaster is 10 ft. above ground
before it starts a certain
portion of the track.
b) Yes. (20, 42) => after 20
seconds the roller coaster
reaches its maximum height
of 42 feet.
c) Yes. (60, 10) => After 60
seconds the roller coaster
returns to a height of 10 feet.
d) No. The height never
equals 0 after the coaster
begins rolling.
e) H(5) = 25.1. After 5
seconds the roller coaster has
reached a height of 25.1 feet
above the ground.
f) At t = 10 seconds and at t =
32.1 seconds. The roller
coaster’s height of 35 feet is
obtained twice during the ride
– once on the way up and
once again on its way down.
At t = 77 the height is again
35 feet, but the roller coaster
ride lasts 60 seconds. 77
seconds is not part of the
domain.
g) After 60 seconds the
height of the coaster is
continually increasing to ∞.
~ page 18 ~
Intermediate Algebra (B)
Unit 6 – Selected Answers
SECTION 6.1B
20) a) 4 mg
b) 2 days
c) Domain: 0 ≤ ≤ 2 after 2 days the drug is completely out of the patient’s bloodstream (y = 0)
Range: 0 ≤ D() ≤ 8. 8mg is the original amount of the drug in the patient’s bloodstream which lessens
each hour afterwards. (Explanations for 20c may vary)
SECTION 6.2A
1) C
7) a) True
10) 4 E
2) A
b) False: 6 + 9 = 15 11) 10L ;
E
G
H
0E
0
3) B
c) False: (2 F ) = 8 F
12) 24 0K
4) B
8) True: Using the power of a
13) 9 M (0N
5) A
power rule, I = I . By the
14) 16;0K
6) B
commutative property of
15) 8 O
multiplication, ab = ba.
9) False: using power of a power
rule ( )H = ∙H = K
SECTION 6.2B
2) 7x + 1
1) Jamal added −3 and − 5,
16) D = 6 − 12 + 9
which are not like terms. The
3) 2<
+ 5< − 4
17) − 2 + 13
correct answer is
18) x + y
4) 10 − 7
H
19) 2x + 6y
9 − 3 − 5 − 2
5) 2 H
Precious also added unlike
20) G + H − 10 + 3
6) 7: − 5: − 3:
terms 5 and − 4. She had a
21) 5 − 10 meters
7) 2 − 9 + 3
second error when she added
22) S(D) = 0.5<
+ 78< + 12,650
8) 10 − 5( + 10
H
H
K
−3 + 8 toget5 . Those
23) −11 + 10
9) 6;
+ 8; − 28
exponents don’t combine. The
24) 14 + 6 + 8
10) 7 + 2 − 3
correct answer is
25) a) 2
11) a + 7b
H
5 + 5 + 4 − 5.
b) 5x
12) 5x + 12z
Kiarra used the distributive
c) 8
13) p – 8q
property incorrectly.
d) 7x
14) 7: − 2: + 2
H
– (– 8 + 3 − 7) = 8 − 3 + 7.
e) –1
15) −6 + 6 ( + 3(
The correct answer is
12 − 12 + 7
1) Explanations may vary
Thao used the product property
of powers incorrectly. He
multiplied and −3 H to get
−3 K . It should be −3 E.
Similarly, he multiplied and
5 to get 5 . It should be 5 H .
The correct answer is
−3 E + 26 H − 4 − 35 + 28
Louis mixed up the +/– symbols
when combining 7x and –3x to
get –10x instead of 4x. The
correct answer is + 4 − 21.
Answers
Intermediate Algebra (B)
SECTION 6.2C
Monique didn’t distribute
the –3 over the last two terms.
The correct answer is
−6 + 15 − 21.
2) 4 H − 20 + 32
3) −7 E + 13 H − 20 4) 3 G − 27 H + 47 − 21 − 98
5) H + 7 − 5 − 35
6) 6 H − 8 − 12 + 16
7) 3 H + 2 − 51 + 70
8) 36 + 60 + 25
9) 5 H − 34 + 99 − 60
10) 6 K − 2 E + 24 G − 20 H
11)
12)
8 E − 6 G + 25 H − 12 + 3 − 18
( = 3 E + 9 G + 2 H + 6 − 8 − 24
y-int. = –24
13) ( = 9 − 42 + 49
y-int. = 49
14) ( = −24 H + 180 + 6 − 45
y-int. = –45
15) ( = G − 4 H + 12 − 4 + 11
y-int. = 11
16) ( = − H − 3 + 8 + 24
y-int. = 24
17) ( = 5 H − 38 + 76 − 33
y-int. = –33
~ page 19 ~
Unit 6 – Selected Answers
Intermediate Algebra (B)
18) Explanations may vary
a) The end behavior is
determined by looking at
the degree (the highest
exponent, which should
be the first term) and the
leading coefficient.
b) The y-intercept is the
constant, or last term in
standard form.
19) Explanations may vary
a) 78.45; In 1995 the average
amount of bananas (in
pounds) eaten per person in
the US was 78.5.
b) Increasing pounds of
bananas consumed per
person from 1995 to 1996,
then decreasing from 1996
to 2000.
SECTION 6.2C
20) a)
3
(continued)
27)
2
3 ( n − 7 ) + 30
R ( t ) = 0.427t + 5.445t + 346.517t + 3679.92
b) $3,679,920,000
c) y-intercept
21) Width = 15 feet
Length = 30 feet
22) a)
3
3n − 21 + 30
3
3 ( n − 7 + 10 )
−n=
−n=
3
n − 7 + 10 − n = 3
R ( t ) = − x3 + 40 x 2 + 6000 x − 240, 000
b) A(0) = $ − 240,000
c) Yes; negative number for
revenue means the costs
were higher than the sales
(which was $0)
(explanations may vary)
23) T = 6 − 2 − 20
24) 20 H − 28 + 15 − 21
25) 10 − 3 − 27
26) T = 53 + 16 − 3
−n=3
3=3
28) a) 3
b) 2x
c) –7
d) 3
29) H + 3 + 3 + 1
30) H + 3 + 2
31) Explanations may vary.
The number of bags of various
flavored chips times the price
per bag gives the amount of
money collected from the
sales.
SECTION 6.2D
3
2
1) 2 x − 3 x − 3 x + 7
Yes, (x + 9) is a factor.
2)
3)
4)
5)
6)
7)
8)
x + 2x −1
Yes, (x – 5) is a factor.
10) #7 and #9; the divisor is a
3
5 x 4 + 7 x3 + 16 x 2 + 2 x − 3 −
factor of the dividend.
x−2
11) Lisa is correct, except to write
No, (x – 2) is not a factor.
the quotient:
31
2x − 6 +
5
2x + 6
x2 + 4 x + 2 −
x−2
No, (2x + 6) is not a factor.
Maut
is
missing
a place
2
−4 x 5 − 4 x 4 − 4 x 3 − 9 x 2 − 6 x − 5 +
holder for the linear (x) term.
x −1
The correct quotient for his
No, (x – 1) is not a factor.
problem:
x2 − 2 x + 2
10 x 3 − 5 x 2 + 9 x − 9
Yes, (2x – 1) is a factor.
Craig is used “–2” for the
12
x2 + 4 x − 1 +
outside value and should have
x+2
used a value of “2” (the value
No, (x + 2) is not a factor.
that creates a zero value for
4 x 2 + 17 x + 16
the divisor. The correct
Yes, (x – 8) is a factor.
quotient for his problem:
4
4
3
2
1
3x + 3 x + 7 x + 7 x + 6 +
x−2−
x −1
x−2
No, (x – 1) is not a factor.
Answers
Intermediate Algebra (B)
9)
2
12)
(x
2
)
− 5 is the length of the
rectangular garden.
13)
( 2x
2
)
+ 4 x − 6 is the base of
the triangle.
14) ( x − 4 ) is the width of the
mural.
15) a) ( x + 2 )
b) B is the divisor, located in
the denominator of the
remainder fraction.
c) A = 5 x 3 − 3 x 2 + 21x − 8
16) One example would be
( 2x
3
)
− x + 3 ÷ ( x − 1)
17) One example would be:
(x
3
) (
− 3x 2 + x − 3 ÷ x 2 − 1
)
Synthetic division can only
be used when the divisor is a
binomial with degree 1, like
( x − c)
~ page 20 ~
Unit 6 – Selected Answers
Intermediate Algebra (B)
1)
a) There are 3 distinct linear
factors, which yield 3
x-intercepts at
( −12, 0 )( −5, 0 ) ,
and ( 8, 0 ) .
b) There is one duplicate
linear factor which gives a
double root at x = –5 (but
only one x-intercept here)
and the other linear factor
yields the x-intercept at
(1, 0 ) .
c) There are 3 distinct linear
factors which yield 3
x-intercepts at
 1 
 − , 0  ( 6, 0 ) , and (10, 0 )
 2 
2)
3)
zero, root, solution
algebraically: set the
function equal to zero (0)
and solve for x.
SECTION 6.2E
graphically: look for the
x-intercept(s) and/or use a
graphing calculator to
“calculate the zeros”.
4) No; there is a remainder
value different than zero.
5) No; there is a remainder of 2.
If (x – 1) was a factor,
there would be a zero
value for a remainder.
6) Yes; because the remainder
is zero, this means that
(x + 4) is a factor of the
polynomial, which means
x = –4 is a zero (or
solution).
7) No; the remainder is 5 (not
zero) so x = 3 is not a
zero.
8) k = –4
9) k = 0
10) f ( x ) = 2 x ( x − 2 )( x + 5 )
11) P ( x ) = x 3 −6 x 2 − 4 x + 24
12) a) y = ( x − 5 )( x + 2 )( x + 1)
b) y = ( x + 1)( x − 5 )( x − 5 )
c) y = ( x + 2 )
d)
13) a)
b)
c)
2
( x − 1)
y = ( x + 1)( x + 5 )( x − 5 )
y = ( x − 2 )( x + 1)( x + 3 )
y = ( x + 4 )( x − 3 )( x − 5 )
y = ( 9 x + 1)( x − 1)( x + 2 )
d) y = ( 2 x + 1)( x + 5 )( x − 4 )
e) y = ( 3 x − 2 )( x + 5 )( x + 2 )
f) y = ( 5 x + 1)( x − 1)( x + 3 )
14) a) Height:
Width:
b) length:
height:
( 3x − 10 )
( x + 1)
( 2x + 5)
( x + 5)
SECTION 6.3A
1) a) Real zeros: x = –10, –1, 3
Factors: ( x + 10 ) ( x + 1)( x − 3 )
2)
1
20
Possible equation:
y=−
1
5
c)
( x + 10 ) ( x + 1)( x − 3 )
b) Real zeros:
x = –3(double root), 2
Factors: ( x + 3 )( x + 3 )( x − 2 )
( x + 3 ) ( x + 3 )( x − 2 )
4)
b) ( −4.879, 0 ) , (1.287, 0 ) , (1.592, 0 )
Possible equation:
y=
a) (1, 0 ) , ( 3, 0 ) , ( 4, 0 )
d)
3)
( 0.438, 0 ) , ( 4.562, 0 ) , ( 9, 0 )
( −3, 0 )
a) x =
1
3
5)
6)
7)
Height: 8 feet
Width: 15 feet
Length: 40 feet
Height: 8.85 feet
Width: 6.85 feet
Length: 13.85 feet
They will need to sell 30 cars
x = 0 meters or x = 6 meters
b) x = −2.059, −0.469, or 0.777
c) x = −2,1, or 3
d) x = −1, or 2
c) Real zeros:
x = –7, –2, 1
Factors: ( x + 7 )( x + 2 )( x − 1)
Possible equation:
y=
1)
1
5
( x + 7 )( x + 2 )( x − 1)
a) Yes; Jebediah knows he
can find the x-intercepts
(solutions) where the
y-value equals zero.
Kalani knows she can look
at the intersection(s) of her
2 graphs to find the
Answers
Intermediate Algebra (B)
SECTION 6.3B
x-values (solutions) where
y = 8.
b) Yes; both are effectively
solving a system of two
equivalent equations by
graphing.
3
2
 y = −0.5 x + 2.5 x + 0.5 x − 2.5
Jebediah  1
 y2 = 0
 y = −0.5 x 3 + 2.5 x 2 + 0.5 x + 5.5
Kalani  1
 y2 = 8
2)
3)
x = –3, –1, 1
x = –3, –1.5, 3
~ page 21 ~
Unit 6 – Selected Answers
Intermediate Algebra (B)
4) x = 2
5) a) x = −2.607,1.392, or 7.715
b) x = −2.574, 0.177, or 4.398
c) x = 4.927
d) x = −7.786, −1.410 or 6.196
6) a) x = −2.067 or 1.284
b) x = −1.446, 0.177, or 2.086
c) x = −4.511, 0, 0.759 or 1.753
d) x = −6.169, −3.623, −1.230 or 7.022
7) There are 3 situations that
demonstrate there are a
total of 3 solutions:
1real solution, 2 complex
solutions (show a graph)
2 real solutions (1 of these
being a double root), 0
complex solutions (show a
graph)
3 real solutions , 0
complex(show a graph)
1)
a)
2)
b)
c)
a)
b)
c)
3)
a)
b)
c)
Answers
x = −2,
−5 ± 3 3
2
x = −5, 3 ± 3i
8)
x = 3, −1, −2 ± 2
Intermediate Algebra (B)
11) a) In the year 2034 and again
in the year 2108
b) 189.3 million; this is a
relative maximum for time
after 1970.
c) When x = 104.3 years after
1970, which is in the year
2074
12) 1989
13) a) ( −2, 6 ) , ( −1, 5 ) , (1, −3 )
b) method 1: using the
graphing method as shown
here, the solutions are the
x-coordinates of the
point(s) of intersection, so
x = –2, –1, 1
method 2: solve this
algebraically. Using
substitution, set the two
expressions equal to each
other:
b) In the year 1996 (1.7 years
after 1995).
c) The average number of
pounds equals 13.5 pounds
three different times:
When x ≈ 1.2 ⇒ year 1996
When x ≈ 2.7 ⇒ year 1997
When x ≈ 5.3 ⇒ year 2000
9) a) 6.3%
b) 26%
c) 47%
10) a) 361 arrests; it is the
y-intercept
b) 14.2 years after, so in the
year 2004
c) In the year 2010
d) About 22 years after 1990,
so in the year 2012; this is
represented by the
x-intercept
SECTION 6.3C
4)
x = 0, 5, −1 ± 11
x = 5,1 ± 33
1
x = − , −3 ± 2 3
3
x = 0, −3, −3 ± 3i
1 5
x = 2, , −
3 2
x = −1, −3 ± 7
a) Graph:
5)
6)
a)
x = −2,
5 ± 17
b)
4
x = 3, −1 ± 2i
c)
x=−
a)
x = 2,
b)
c)
a)
b)
c)
d)
e)
f)
1 −1 ± i 3
,
2
2
−3 ± 5
2
x = −2, 4,1 ± i 7
x = 3 (double root), −2 ± 3
x = 5, −2, 0.75
x = −4,1.5,10
x = 2, −1 ± i 3
x = −3, 0, 7
x = −2,1, 2
x = 3, ± i 3
y1 = y2
3
2
x + x − 5x = − x2 − 4 x + 2
x3 + 2 x 2 − x − 2 = 0
( x + 1)( x − 1)( x + 2 ) = 0
x = −1, x = 1, x = −2
7)
A cubic function always has 3
solutions:
Either 1 real and 2
imaginary solutions (show
graph with 1 x-intercept)
2 distinct real solutions
(with one of them being a
double root – creating three
solutions in total) (show graph
with curve that crosses over
the x-axis and one section that
has a vertex that touches the
x-axis).
3 real solutions (all
different) (show graph with
curve crossing over the x-axis
three distinct times).
~ page 22 ~
Intermediate Algebra (B)
Unit 6 – Selected Answers
Unit 6 – Review Material
1) a) Standard Form
b) It is the y-intercept
c)
x
y
–3
–20
–2
0
–1
6
0
4
1
0
2
0
3
10
d) Rel min: none
Rel max: none
Domain: all real numbers
Range: all real numbers
Inc. int.: −∞ < x < ∞ , or ℝ
Decr. int.: none
x-intercept: ( −1, 0 )
d) Rel min: ( −1.22, −2.11)
Rel max: ( 0.55, 0.63 )
Domain: all real numbers
Range: all real numbers
Inc. int.: −1.22 < x < 0.55
Decr. int.: x < −1.22, x > 0.55
Zero(s): ( 0, 0 ) , (1, 0 ) , ( −2, 0 )
y-intercept:
d) Rel min: (1.54, −0.88 )
Rel max: ( −0.87, 6.06 )
Domain: all real numbers
Range: all real numbers
Inc. int.: x < −0.87, x > 1.54
Decr. int.: −0.87 < x < 1.54
Zero(s): ( −2, 0 ) , (1, 0 ) , ( 2, 0 )
2) a)
2
y = −3 x + 12 x − 7
b) y = x 3 + 2 x 2 − 43 x − 140
3) a) Intercept form or Factored
form
b) It brings forward the
information for finding the
x-intercept
c)
x
y
–3
12
–2
0
–1
–2
0
0
1
0
2
–8
3
–30
Answers
Intermediate Algebra (B)
( 0, 3.5 )
y-intercept:
( 0, 0 )
4) a) It is similar to the vertex
form of a quadratic function,
but this is a cubic function.
b) It provides information about
the inflection point.
c)
x
y
–2
–9.5
–1
0
0
3.5
1
4
2
4.5
3
8
5) a)
b)
c)
d)
e)
f)
3
2
14 x + 2 x +3 x − 6
5 x 3 − 10 x 2 + 18
2 x2 + 5 x + 3
3 x 3 + 16 x 2 − 17 x − 30
2 x 3 + 24 x 2 + 96 x + 123
x2 − 6x + 3
g) x3 − 5 x 2 + 18 x − 36 +
6) a)
b)
c)
d)
e)
f)
7) a)
70
x+2
During month #5
100 coats
50 coats
x > 5.5
0 < x < 5.5
no
f ( x ) = ( x + 1)( x − 1)( x − 7 )
b) f ( x ) = ( x − 1)( x − 4 )( x + 3 )
8) a) x = −5, −3, 3
4
b) x = −5, − , 2
3
1
1
2
3
c) x = 0, , −
d) x = −3, −2,1
[8(a) – 8(d) the procedure for
finding the rational zeros may
vary.]
1
9) a) x = , 4, −3
2
1
b) x = , −6, 2
2
10)
x = −3, 5,
−1 ± i 3
2
11) a) x = −2, −1,
b) x = 3, 4,
−1 ± i 11
2
3± 5
2
~ page 23 ~
Intermediate Algebra (B)
Unit 7 – Selected Answers
SECTION 7.1A
1. a)
•
b)
From the store, drive
South on Aspen St
• Turn Right on Elm St
• Turn Left on Acorn St
• Go past one stop sign and
turn Right/West on Hwy 27
• Travel 8mi to your home
b) I drew a picture of the route
TO the store and then followed
it backwards.
c) Reverse
2. a) 1
b) 15
g) i)
x
P(x)
x
N(x)
1
5
5
1
2
10
10
2
3
15
15
3
4
20
20
4
ii) They are reversed.
4. a)
b)
x
f -1(x)
x
f(x)
–3
4
4
–3
1
–2
–2
1
5
2
2
5
c)
(VWH)X0
E
EVW0
= + 3
e) An inverse function is the
reverse process of a
function; it “undoes” what a
function “does.”
3. a)
Pairs of Skates Additional Pay
4
$20
$100
20
8
$40
$65
13
x
$5x
Z
$x
[
b) The extra pay P(x) is 5
times the # skate pairs
sharpened x.
c) The number of pairs of
skates N(x)
0
is
E
the extra
pay x.
d) P(3) = 15; If he sharpens 3
pairs of skates, he will
make $15 additional pay.
e) N(15) = 3; If he made $15
additional pay, he must
have sharpened 3 pairs of
skates.
f) They give the same results,
backward.
Answers
Intermediate Algebra (B)
8.
x
–3
–2
–1
0
1
2
3
y
c)U() = d)Y()
c) Yes – Each input (x) has
only one output (y).
5
-5
5 x
y = 2x
–6
–4
–2
0
2
4
6
0
y= x
–6
–4
–2
0
2
4
6
–3
–2
–1
0
1
2
3
y
-5
5
d) No – Each input (x) does
not have exactly one
output (y).
[ex. (0, 3) and (0, –0.5)]
5. a) {(5, –13), (–9, –9), (–4, 0), (6, 4),
-5
5 x
(10, 9)}
b) Yes – Each input (x) has
only one output (y).
6. a) {(4, –2),(7, 4),(11, 0),(7, –3)}
b) No – the input 7 has two
different outputs (4 and –3)
7. a)
x
–2 –1
0
1
2
g-1(x) –10
–6
–2
2
6
-5
Yes – the inputs (x) and
outputs (y) are reversed.
9.
x
0
1
2
3
4
5
6
7
y = x2
0
1
4
9
16
25
36
49
x
0
1
4
9
16
25
36
49
~ page 24 ~
y = √
0
1
2
3
4
5
6
7
Intermediate Algebra (B)
Unit 7 – Selected Answers
brought with us, given we
received a known amount
of British pounds.
\
c) 4 =
y
5
4
3
0.EG
2
d) $191.56
11. -Choose points on the graph.
-Reverse the coordinates
(x, y) (y, x)
-Graph the new coordinates
1
-4
-3
-2
-1
1
2
3
4 x
1
2
3
4 x
-1
-2
-3
-4
-5
Yes – the inputs (x) and
outputs (y) are reversed.
10. a) 770 pounds
b) It would tell us how much
money (in US dollars) we
SECTION 7.1B
1. a) Values lower than 3 would
make a negative in the √
b) The smallest √ result would
be 0, and then adding 5
gives a minimum value of 5,
so y cannot be less than 5.
c) Think about (guess/check)
what #s can’t work for x and
then #s that won’t be results
for y. (see a) and b))
d) There are no points graphed
with an x-value before 3;
there are no points graphed
with a y-value less than 5.
e) There are “error”s before x = 3
and no y-values below y = 5.
2. a) Equation
Domain: x > 7
Range: y < –4
Explanation: an x-value
lower than 7 would make a
neg. under the √; the greatest
√ result is 0, and then
subtracting 4 gives a max
value of –4, so y cannot be
greater than –4.
b) Graph
Domain: x > –8
Range: y > –7
Explanation: There are no
points graphed with an
x-value before –8; there are
no points graphed with a
y-value below 7.
Answers
Intermediate Algebra (B)
2. c) Table of values
Domain: x > 3
Range: y > 5
Explanation: There are
“error”s before x = 3 and no
y-values less than 5.
3. Milo is correct – Basra appears
to be thinking about the
range.
4. [B]
5. [C]
6. [A]
7.
8.
y
5
4
3
2
1
-4
-3
-2
-1
-1
-2
-3
-4
-5
Decreasing
Domain: x > –4
Range: y < 3
x-int: (5, 0) y-int: (0, 1)
9.
y
2
1
-4
-3
-2
-1
1
2
3
-1
Increasing
Domain: x > 0
Range: y > 2
x-int: none y-int: (0, 2)
-2
-3
-4
-5
-6
-7
-8
Decreasing
Domain: x > 0
~ page 25 ~
4 x
Unit 7 – Selected Answers
Intermediate Algebra (B)
Range: y < –2
x-int: none y-int: (0, –2)
11.
13. Shifted Left 3 and Down 4
14. Flipped
Shifted Right 2 and Up 515.
10.
Graph Number 2
Beginning Point (0,0)
x-intercept (0,0)
3
1
4
(-6,0) (2, -4) (-2,4)
(-6,0)
y-intercept (0,0) (0,2.6) None (0,2.6)
Inc./Dec. Dec.
Inc.
Inc.
Dec.
Domain x > 0 x > -6 x > 2 x > -2
Range y < 0 y > 0 y > -4 y < 4
Increasing
Domain: x > 0
Range: y > 0
x-int: (0,0) y-int: (0,0)
Increasing
Domain: x > 1
Range: y > 2
x-int: none y-int: none
12.
16. Domain: x > –3 Range: y > –4
17. Domain: x > 4 Range: y < 1
18. Answers may vary
a) ( √ 5 2
b) ( √ 7 10
c) ( √ 9
Increasing
Domain: x > –5
Range: y > –4
x-int: (–1, 0) y-int: (0, 0.47214)
1. a) No – you can ∛ any number
(pos or neg).
b) No – since x can be anything
^
and √ could be anything,
adding 6 could lead to any #.
c) (–3, 6)
d) The x-coordinate is the
result of setting what is under
the radical equal to zero and
then solving for x; the
y-coordinate is the value of
the constant added/subtracted
outside the ∛.
e) Increasing – it is a positive ∛.
Answers
Intermediate Algebra (B)
SECTION 7.1C
2. a) Increasing
Domain: 7
Range: 7
Point of Inflection: (2, –4)
b) Decreasing
Domain: 7
Range: 7
Point of Inflection: (3, –5)
3. Neither – should be (12, 10)
Arturo didn’t set what is under
the radical equal to zero and
then solving for x.
Kira has the x and y switched.
4.
y
8
4
-8
-4
4
-4
-8
Decreasing
Domain: 7
Range: 7
Point of Inflection: (0, 0)
~ page 26 ~
8 x
Unit 7 – Selected Answers
Intermediate Algebra (B)
7.
5.
-12
-8
y
y
8
8
4
4
-4
4 x
-8
-4
-4
4
10.
11.
12.
13.
[C]
[A]
[B]
a) Answers will vary:
Graph: Identify where the
curve changes from a hill to a
bowl.
Table: Look for a set of three
y-values that are the same
distance apart; the inflection
point is the middle of these
three.
Equation: The x-coord. is
calculated by setting what is
under the radical equal to zero
and then solving for x; the
y-coordinate is the number
added/subtracted outside the ∛.
b) Answers will vary:
Graph: Look left to right to see
if it is going uphill or downhill
Table: Look to see if y-values
increase or decrease (as x-values
increase.)
Equation: Look for the sign on
the number in front of the ∛;
positive is increasing,
negative is decreasing.
14.
15.
16.
17.
Increasing, (7, –4)
Decreasing, (–9, 0)
Shifted Left 2, Down 4
Flipped; Shifted Right 3, Up 4
8 x
-4
-8
-8
Decreasing
Domain: ℝ
Range: ℝ
Point of Inflection: (–3, 5)
Increasing
Domain: ℝ
Range: ℝ
Point of Inflection: (–4, 1)
8.
6.
y
8
y
8
4
4
-12
-8
-4
4
-8
-4
4
x
8 x
-4
-4
-8
-8
Increasing
Domain: ℝ
Range: ℝ
Point of Inflection: (0, 4)
Decreasing
Domain: ℝ
Range: ℝ
Point of Inflection: (–3, –2)
9.
y
8
18.
4
Graph Number
4
1
Point of Inflect. (-2,6) (-1,3)
-12
-8
-4
4
x
y-intercept
3
(5,0)
Inc/Decreasing Dec.
Increasing
Domain: ℝ
Range: ℝ
Point of Inflection: (–5, –1)
Answers
Intermediate Algebra (B)
(0,-4)
(0,4.8) (0,4) (0, 1.8) (0,-4)
Inc.
Dec.
-4
-8
2
19. Answers will vary:
^
a) ( = √ + 3 + 5
^
b) ( = − √ − 4 − 6
^
c) ( = √ − 2 + 9
~ page 27 ~
Inc.
Unit 7 – Selected Answers
Intermediate Algebra (B)
1. [A] 2. [C] 3. [A]
4. [B] 5. [A] 6. [C]
7. a) False: should = 2 ∙ 2 ∙ 2
b) True
c) False: should = g ( H F
d) True
e) False: should = 15 f) False: should =
Mh ij
V ik
8. Katiana
( H )N = W[ E
N = N
9. a) 4
b) –3
c) 0
d) any real number except 0
10. Sarah made an error when
simplifying the denominator in
Line 2: the –2 exponent
should have gone to the p and
d, not the q. The denominator
should simplify to l W0 <G 4WK
and a final answer of p 4 q11
SECTION 7.2A
11. Tyrell also made an error in
Line 2: 2WG should simplify to
0
0K
(not –8). The final answer
0
should be M O .
12. Answers will vary:
-Simplify: (( H )W
to W
( WK
3 x5 x −2 y −6
So:
18 x −4 y 5
H
0M
V^
=
V mn
mp
o
-Simplify y’s: k =
o
Vq
-Final Answer: ii
Ko
-Simplify x’s:
1
16.
16 p 5
c9 p3
r12
17.
18. 9b 6 z12
0
=K
g
1
2k 5
15.
-Simplify x’s in numerator:
3 x 3 y −6
E W
= H So:
18 x −4 y 5
-Reduce numbers:
13. Yes – Variable parts match so
they are “like terms” = 5 H ( G
14. 4
9 x4
2
19.
0
( W00 = oii
SECTION 7.2B
1. [H] 2. [B] 3. [D] 4. [E]
5. [G] 6. [A] 7. [F] 8. [C]
9. [C] 10. [B]
11. No – Exponent is NOT neg, so
exponent should not be moved
to denominator.
Ans = √17
12. Yes
13. Yes
i
k
14. No 5k ≠ 1 . Ans = √5s
15. Yes
n
1. Never: √−256 is not real
2. Always
3. Always
4.
5.
6.
7.
Never: 3 343 = 7
Sometimes
Sometime
Sometimes
Answers
Intermediate Algebra (B)
25. ☺
16. Yes
45
5
4
17. No: 4 = 4
18. [D]
19. [C]
20. ☺
21. (54)0/G or 50/G 40/G
16
22. ( 36 fg ) = 361 6 f 1 6 g1 6 ; ( 361 6 ≠ 6 )
23. ☺
(
24. b 2 + c 2
15
)
; cannot be
simplified since there is no
power of a sum exponent
property.
SECTION 7.2C
#5 - #7: Since the radical on the left
side of the equation has an
even root, the answer must be
greater than or equal to zero.
However, the right side of the
equation has a variable to an
odd power, which could be
26.
29.
32.
0
H
G
g
M
g|u^ |
27. 5
30.
33.
HGH
0
E
0
0K n
28.
0
0N
G
31. 7d
34. 806.76≈ 806 people
35. 13,770mm
36. True: −360/
= −1 ∙ 360/
= −6
n
37. False: √−16 is not real
either a positive or a negative
answer.
8. Always
9. 3
10. 16
11. 9
12. 8
13. 25
14. 0.3
15. 0.6 G
18. 64YG ℎ
16.
0N
H
17. 6v4E/H
~ page 28 ~
Unit 7 – Selected Answers
Intermediate Algebra (B)
19. 343|4
20.
21.
E
G
KG
xp
H|
22. [A], [C], and [D] require
absolute value symbols
around one of the variables.
23. ≈ 1.34 in
24. ≈ 1.68 ft
ANSWER KEY
1. 2
3.
0
O
2. 2.520
≈ 0.111
5. 16
4.
1
1
=
2
49
2401
7. √6 ≈ 2.449
8.
0N
K
E
≈ 0.016
9. 100
10. 6b
Quiz grade: 2/10
Incorrect: (1, 3, 4, 6, 7, 8, 9, 10)
11. Neither Jerome nor Cambria
have correct answers:
1. a) Intersection at (16, 4); x = 16
b) Intersections at (–2, 0) and
(–1, 1); x = –2 or x= –1
c) Intersections at (–3, 4) and
(13, 4); x = –3 or x=13
d) No intersections;
No Real Solutions
e) Intersection at (2,√2); x = 2
f) Intersection at (31.5, 5);
x = 31.5
1. x = 9; Yes, one intersection on
graph at x = 9
2. x = 5; Yes, one intersection on
graph at x = 5
3. x = 5; NO, graphs do not
intersect so NO REAL
SOLUTION
4. x = 5 or x = 2; Yes, two ints on
graph at x = 5 and x = 2
Answers
Intermediate Algebra (B)
≈ 0.48 cm
234
cm
864 cm2
≈ 0.165 yz 16.5%
SECTION 7.2D
1
−1 6
( −64 ) =
16
( −64 )
19. 4
20. 12a 53 12
which is not a real number.
21.
(−64)K = 68,719,476,736
12. ?
13. ?
6. 1.5
25.
26.
27.
28.
0
=
0
=
H
14. no number possible to replace
the question mark
15. ? =
0
H
H
0
=−
16. ? =−
17. ?
0
\ji/ip
22. k 19 6
23. z
24. |s|
25. 2
26. H/
27. 6
28. 9
29. 1
30. Over 162 trillion years
18. 36
2.
3.
4.
5.
SECTION 7.3A
a) x = 6
b) x = 36
c) x =0.807 or x =6.193
d) x = 2.5 e) x = 1
f) x = 11 or x = –16
g) x = 2 or x = 18
h) No Real Solution
i) x = –39
4 ≈ 72.727 ft
| ≈ 149.34 million km
a) @ ≈ 261.2 km/hr
b) 4 ≈ 0.108 km
SECTION 7.3B
5. x = 7 or x = -3; No, only one
intersection at x = 7 so x = 3 is
extraneous.
6. x = 2; Yes, one intersection on
graph at x = 2
7. x = 10; Yes, one intersection on
graph at x = 10
8. x = 5 or x = –4; Yes, two ints on
graph at x = 5 and x = –4
9. For problem #3, when x = 5, the
‘check’ yields 8 = 2. This is a
6. a) @ ≈ 1.924 sec
b) } ≈ 5.066 ft
7. No – If the graphing calculator
yields decimal solutions that do
not terminate or repeat, the
solutions (irrational) can be
found algebraically.
false sentence therefore the
value of 5 is not a solution to
the equation.
For problem #5, when x = –3,
the ‘check’ yields –6 = 6. This
is a false sentence therefore
the value of –3 is not a
solution to the equation.
~ page 29 ~
Unit 7 – Selected Answers
Intermediate Algebra (B)
10. a) Graphically: The graphs do
not intersect for the
extraneous x-value.
b) Algebraically: A solution
comes out of the work but
doesn’t check so is not a
true solution.
11. Serge did the problem
correctly. Jeremy needed to
subtract 25 BEFORE squaring
both sides
12. Indy did the problem correctly.
Latishia did the work correctly
but did not check for
extraneous solutions; x = 3 is
extraneous. Sango squared
(2 – x) incorrectly;
(2 – x)2 = (2 – x) (2 – x)
not 22 – x2
13. x = 5
14. n = 77
15. No real solution
16. No real solution
17. x = 7 18. x = –13
19. x = –8 20. x = 1 21. r = 8
22. No real solution
H
23. x = E
24.
25.
26.
27.
x = ±32
d = 2 ± 2√2
b = 20
x = 621
28 – 33: Answers may vary
28. graphically
29. algebraically
30. graphically
31. algebraically
32. algebraically
33. algebraically
34. Answers will vary:
√5 + 3 = −2
35. ℎ ≈ 0.6;
36. a) f = 30 gallons/min
b) p = 640,000 lbs/in2
37. t ≈ 4.15 ; It would take
4.15 hours.
38. x = 58.81 mi2
Unit 7 - Review
1. a)
-2
c)
y
y
6
5
6
5
4
3
2
4
3
2
1
1
-1
2
4
6
8
10 x
-2
-1
-2
-3
-4
-2
-3
-4
-5
-6
-5
-6
Domain: x > 0
Range: y > 0
b)
e)
y
6
5
4
3
2
2
4
6
8
10 x
1
-2
-1
2
4
6
8
10 x
6
8
10 x
-2
-3
-4
-5
-6
Domain: x > 0
Range: y > –2
Domain: x > 0
Range: y < 0
y
f)
6
5
d)
4
3
2
-1
2
4
6
8
4
3
2
3
2
10 x
-2
-3
-4
1
1
-2
-5
-6
-1
2
4
-2
-3
-4
Domain: x > –1
Range: y > 0
Answers
6
5
6
5
4
1
-2
y
y
Intermediate Algebra (B)
-5
-6
Domain: x > –2
Range: y > –3
6
8
10 x
-2
-1
2
4
-2
-3
-4
-5
-6
Domain: x > –1
Range: y < 2
~ page 30 ~
Unit 7 – Selected Answers
Intermediate Algebra (B)
2. Answers may vary:
( √ 2 + 6
3. a)
d)
5
y
4
5
3
4
2
3
1
2
-5 -4 -3 -2 -1
1
2
3
4
-1
EN
1
2
3
4
5 x
-4
-3
-5
^
9. √ 0N
-4
-5
Domain: 7
Range: 7
Inflection Point: (0, 0)
b)
e)
13. 6 G
5
5
4
4
3
3
2
2
1
1
1
2
3
4
5 x
-5 -4 -3 -2 -1
11. 27 0E g
14.
-2
-3
-4
-4
-5
-5
M
2
3
4
5 x
0
19. 38v G 4
ft
20. 104,573 people
M
21. = H
22. x = −1 or
c)
x = −2
23. x = 7
f)
24. x = 5 or x = –1
5
y
4
5
3
4
2
3
1
-5 -4 -3 -2 -1
17. a
18. − E
Domain: 7
Range: 7
Inflection Point: (0, 0)
y
Vh n/^
oj
16.
gV^
1
-3
0
g
15. 16 E
-1
-2
Domain: 7
Range: 7
Inflection Point: (–1, 0)
G~k
Ep
12.
y
-1
10.
Domain: 7
Range: 7
Inflection Point: (1, 2)
y
-5 -4 -3 -2 -1
7. 343
8. 5i
-3
-2
0
6. V k
-2
5 x
0
GO
5. 0
EV ^
-1
1
-5 -4 -3 -2 -1
4.
y
2
1
2
3
-1
4
1
5 x
-2
-5 -4 -3 -2 -1
1
2
3
-1
-3
-2
-4
-3
-5
-4
4
5 x
-5
Domain: 7
Range: 7
Inflection Point: (0, 2)
Answers
Intermediate Algebra (B)
Domain: 7
Range: 7
Inflection Point: (2, 1)
~ page 31 ~
Unit 8 – Selected Answers
Intermediate Algebra (B)
SECTION 8.1A
9.
1. Opens: Up
Axis: x = –2
Vertex: (–2, –6)
2. Opens: Down
Axis: x = 3
Vertex: (3, 4)
3. Opens: Up
Axis: x = 5
Vertex: (5, –4)
4. Opens: Down
Axis: x = 2
Vertex: (2, –5)
5. Opens: Down
12.
y
8
8
4
4
0
0
x
-4
x
-4
-8
-8
-4
0
4
-8
-8
8
Domain: x = ℝ
Range: y > 4
0
0
10.
Vertex: (H, –4)
6. Opens: Up
Axis: x = 2
Vertex: (2, 4)
7.
0
-4
4
-8
-8
0
x
4
-4
0
4
8
11.
Domain: x = ℝ
Range: y > –4
4
8
0
x
4
-4
0
x
-4
0
4
8
Domain: x = ℝ
Range: y > 0
Answers
-8
-8
-4
0
Domain: x = ℝ
Range: y < 3
-4
Intermediate Algebra (B)
Vertex: (2, 0)
Opens: Up
Domain: x = ℝ
Range: y > 0
Steepness: m = ±3
Vertex: (6, 0)
Opens: Up
Domain: x = ℝ
Range: y > 0
Steepness: m = ±3
y
8
y
15.
8
Domain: x = ℝ
Range: y < –3
-4
-8
-8
8
14.
x
8.
4
13. Domain: x = ℝ
Range : y > 3
y
4
8
0
0
8
y
-4
-4
Domain: x = ℝ
Range: y < 5
Axis: x = H
-8
-8
y
4
8
16. Both have a –6 in the absolute
value bars with the x.
17. #14 has the multiplier of 3
inside the absolute values bars;
#15 has the multiplier outside
the absolute value bars.
18. When the 3 is on the inside, it
influences the x-coordinate of
the vertex by dividing the 6 by a
factor of 3.
19. [equation B]
20. [D]
21. y = 2x + 8, x > –4
y = –2x – 8, x < –4
~ page 32 ~
Unit 8 – Selected Answers
Intermediate Algebra (B)
SECTION 8.1B
1.
5.
9.
2.
6.
10.
3.
7.
11.
4.
8.
12.
Answers
Intermediate Algebra (B)
~ page 33 ~
Unit 8 – Selected Answers
Intermediate Algebra (B)
13.
14.
15. [B] and [D]
16. Answers may vary.
a. ( / | 2| 5
b. ( > | − 4| 3
c. ( / | 2| 6
17. [C]
18. [B]
19. [A]
20. [D]
Domain: x = ℝ
Range: y = ℝ
Domain: x = ℝ
Range: y > 0
SECTION 8.2A
3. Graphical Solution
1. Graphical Solution
y
5. Graphical Solution
y
4
2
0
x
y
8
8
6
6
4
4
2
2
0
x
-2
-4
-2
0
2
-2
-10 -8 -6 -4 -2 0 2 4
4
Algebraic Solution
x = 3 x = –3
2. Graphical Solution
y
0
x
-2
-10 -8 -6 -4 -2 0
Algebraic Solution
x = 2 x = –8
Algebraic Solution
–5 < x < –1
4. Graphical Solution
6. Graphical Solution
y
y
10
10
10
8
8
8
6
6
6
4
4
4
2
2
2
0
-2
-4
0
x
-2
0
2
4
-2
-4
x
-2
0
2
Algebraic Solution
Algebraic Solution
x = 2 x = –2
Answers
Intermediate Algebra (B)
g
x=H
2 4
x = –1
4
0
-2
-2
x
0
2
4
6
Algebraic Solution
x < 2 or x > 6
~ page 34 ~
8
Unit 8 – Selected Answers
Intermediate Algebra (B)
7. Graphical Solution
9. Graphical Solution
16. {all REAL numbers}
y
y
6
6
17. 2 < x < 6
4
4
2
0
2
x
-2
0
x
-2
-2 0 2
4 6
-4
-6
-4
8 10 12
-2
0
2
Algebraic Solution
Algebraic Solution
x < 6 or x > 10
–2 < x <
8. Graphical Solution
G
H
10. –10 < x < –4
y
8
6
4
18. a) Answers will vary:
16, 16.1, 16.6, 15.9, 15.4
b)
c) 15.4 < x < 16.6
d)| − 16| ≤ 0.6
19. a) Answers will vary:
40℉, 41℉, 42℉, 74℉, 75℉
b)
11. x > 9 or x < –1
4
2
12.
0
W00
K
//
g
K
x
-2
-4
-2
0
2
4
Algebraic Solution
–1 < x < 0
13.
W
H
c) x < 43 or x > 73
d) | − 58| > 15
20. a) Answers will vary:
3, 4, 4.5, 5, 6
b)
<x<4
14. –4 < x < 8
c) 3 < x < 6
d) | − 4.5| ≤ 1.5
15. x < –6 or x > 10
Answers
Intermediate Algebra (B)
~ page 35 ~
Unit 8 – Selected Answers
Intermediate Algebra (B)
Unit 8 - Review
1. a)
b)
c)
d)
3.
Up
(3, –1)
x=3
±2
8. x = 8
x=0
9. x > 2 or x < –7
y
8
10. –9 < x < 3
6
4
11. a) 230 < x < 270
b) | − 250| ≤ 20
2
4.
y
x
-2
-2
2
4
6
12. a) 7.7 < x < 8.3
b) | − 8| < 0.3
6
8
4
2
2. a) Down
b) (–1, 4)
c) x = –1
0
d) ± H
-4
y
-6
-6 -4 -2
8
6
2
4
6
5. Answers may vary:
( = 4| − 1| − 3
4
2
x
-2
-6 -4 -2
Answers
x
-2
2
4
6
Intermediate Algebra (B)
6. Answers may vary:
−1
| + 2| + 1
(=
3
7. x =
N
H
x=
W0N
H
~ page 36 ~