MATH 105 Test 2 (Sec: 1.6 - 4.3) Version 0
Sec Number:
Name: KEY
1. (10 pts) A 70-ft by 60-ft parking lot is torn up to install a sidewalk of uniform width around
its perimeter. The new area of the parking lot is two-thirds of the old area. How wide is the
sidewalk?
Let x be the width of the side walk.
Originally the area of the parking lot was:
area = 70(60) = 4200 square feet
The new area is
new area = (70 − 2x)(60 − 2x) = 2800.0 square feet
Thus,
2800.0 = (70 − 2x)(60 − 2x)
=⇒ 4x2 + (−260)x + 1400.0 = 0
Use the quadratic formula:
√
−b ± b2 − 4ac
=⇒ x = 59.0753645318366, and x = 5.92463546816338
x=
2a
Here we reject the larger answer as it is outside the domain of physical answers. The width
of the sidewalk is approximately 5.925 feet wide.
2. (10 pts) A ball is thrown directly upward from a height of 4ft with an initial velocity of 34ft/s.
The function
s(t) = −16t2 + 34t + 4
gives the height of the ball, in feet, t seconds after it has been thrown. Determine when the
ball reaches its maximum height. What is the balls maximum height? At what time does the
ball hit the ground?
The maximum height of the ball is found at the vertex of the given model:
−34
17
17
353
vertex =
,s
=⇒ s
=
≈ 22.063
2(−16)
16
16
16
Thus the maximum height of the ball is 22.063 feet, and this happens
17
16
after the ball is thrown. The ball hits the ground at the positive root value.
1
−b − ((b2 ) − 4ac) 2
t=
= 2.2368
2a
The ball hits the ground after 2.2368 seconds.
≈ 1.063 seconds
3. (8 pts) Solve.
4x
6
7
+ = 2
x+7 x
x + 7x
Multiply both sides by x(x + 7) and arrive at:
4x2 + 6(x + 7) = 7 =⇒ 4x2 + 6x + 35 = 0
p
−6 ± 36 − 4(4)(35)
=⇒ x =
2(4)
=⇒ x = −0.750000000000000 + 2.86138078556490i
and x = −0.750000000000000 − 2.86138078556490i
Here each value if it is not complex should be checked to see if it works in the expression.
4. (4 pts) Solve:
x4 + 13x2 = −42
Let y = x2 then
y 2 + 13y + 42 = 0 =⇒ (y + 6)(y + 7) = 0
y = −6 and y = −7
=⇒ x2 = −6 and x2 = −7
Thus,
√
√
x = ± −6 and x = ± −7
Note that some of the roots may be imaginary or complex.
5. (8 pts) Solve.
√
√
Check:
x−
x−
√
x−8=5
√
√
√
x − 8 = 5 =⇒
x= x−8+5
√
=⇒ x = (x − 8) + 10 x − 8 + 25
√
=⇒ 8 − 25 = 10 x − 8
17 √
=⇒ − = x − 8
10
289
=⇒
=x−8
100
1089
=⇒ x =
≈ 10.89
100
√
10.89 −
√
10.89 − 8 = 1.6
6. (10 pts) Use polynomial long division and the fact that x = 9 is a root of
f (x) = x3 − 16 x2 + 73 x − 90
to fully factor f (x). State all three roots of f (x). Polynomial division shows that c is a root.
x2 − 7x + 10
x−9
x3 − 16x2 + 73x − 90
− x3 + 9x2
− 7x2 + 73x
7x2 − 63x
10x − 90
− 10x + 90
0
Thus,
f (x) = (x − 9)(x2 − 7x + 10) = (x − 2)(x − 5)(x − 9)
and the function f (x) has roots at x = 9, x = 2, and x = 5.
− 2 ≤ 6, write interval notation for the solution set, then graph the
7. (6 pts) Solve 7x+8
5
solution set on a number line.
7x + 8 5 ≤8
Now consider the problem in two parts:
32
7x + 8
≤ 8 =⇒ 7x + 8 ≤ 40 =⇒ 7x ≤ 32 =⇒ x ≤
5
7
Or
7x + 8
48
≥ −8 =⇒ 7x + 8 ≥ −40 =⇒ 7x ≥ −48 =⇒ x ≥ −
5
7
Thus,
48 32
x∈ − ,
7 7
8. (4 pts) Solve:
x2 − 1x − 42 = 0
Note that the expression factors as
(x + 6)(x + −7) = 0
Thus,
x = −6 and x = 7