Download MAT098 Distance Test 4

Math 098: Intermediate Algebra
1.
TEST 4: Review Solutions
 121
Write as a simplified imaginary number:
 121  121   1
11i
 11i
2.
 72
Write as a simplified imaginary number:
 72  36 2  1
6i 2
 6i 2
3.
Add and simplify:
 7  8i    6  3i 
 7  8i    6  3i   7  6   81  3i 
 13  11i
4.
Multiply and simplify:
4i  3  2i
13  11i

4i  3  2i
  12i  8i 2
 12i  8 1
8  12 i
 8  12i
5.
Simplify, writing as a complex number:
2   20

4



2   20
4
2  4 5 1
4
2  2i 5
4
2 1 i 5
4
1 i 5
2


1
2
 25 i
 21  25 i
6.
Divide and write as a complex number in the form a + bi:
3  7i
2i
3  7i 3  7i 2  i


2i
2i 2i

6  3i  14 i  7i 2
22  i 2
6  17 i  7 1

4   1
 1  17 i

5
  51  17
i
5
 51  17
i
5
7.
Solve
x  42
 75 using the square root property.
x  42  75
x  42
  75
x  4  5 3 or 4  5 3
x  4  5 3
x  45 3
x  4  5 3 or 4  5 3
8.
Solve x 2  8 x  6  0 by completing the square.
Simplify your answers but do not approximate the solutions.
x 2  8x  6  0
x 2  8x  6
2
8
8
x 2  8x     6   
2
 
2
2
x  4  22
x 2  8 x  16  6  16
or
 4  22
x  42  22
x  42
  22
x  4   22
x  4  22
x  4  22
9.
or
 4  22
3x 2  2x  3  0
Use the quadratic formula to solve:
Simplify your answers but do not approximate the solutions.
Using the Quadratic Formula x 
x
  2 
 b  b2  4ac
with a  3, b  2, and c  3 :
2a
 22  4  3  3
23

2  4  36
6

2   32
6

2  4i 2
6

2  4i 2
6

2 1  2i 2
6

1  2i 2
3

1
3

x
1
3
 2 32 i

2 2
3
 32  16  2   1
 4i 2

i
or
Math 098 Test 4 Review Solutions
1
3

2 2
3
i
Page 2
1
3

2 2
3
i
or
1
3

2 2
3
i
10.
Find all solutions, real and imaginary, of the equation: t 4  7t 2  18
t 4  7t 2  18
t 4  7t 2  18  0
t
2


 9 t2  2  0
2
t 9  0
t2  2  0
or
2
3,  3, i 2 ,  i 2
2
t 9
t  2
t 9
t  2
t  3
t  i 2
t  3,  3, i 2 ,  i 2
11.
Solve for x:
x7  x5

x7  x5
x7
2  x  52
x  7  x 2  10 x  25
x 2  9 x  18  0
x  3x  6  0
x 30
x60
or
x  3
x  6
x  3
Checking for extraneous solutions:
At x  3,
x  7  4  2 and x  5  2 checks
At x  6, x  7  1  1 and  6  5  1 does not check
Therefore the only solution is x  3
12
Solve for x:
3x  1  x  1  2
3x  1  x  1  2
3x  1  2  x  1
 3x  12  2 
2
x 1
2
3x  1  2  2  2 x  1 


x 1
2
3x  1  4  4 x  1  x  1
x  12  2 x  1
x 2  2 x  1  4x  1
2
x 2  2x  1  4 x  4
2x  2  4 x  1
x 2  6x  5  0
x  5x  1  0
x 50
or
x 5
2x  1  4 x  1
2x  1 4 x  1

2
2
x 1 2 x 1
x  1 or x  5
x  1 0
x 1
Checking for extraneous solutions:
At x  5,
3 x  1  x  1  16  4 At x  1,
 42
3x  1  x  1  4  0
 2 checks
 2 checks
Therefore the solution is x  1 or x  5
Math 098 Test 4 Review Solutions
Page 3
13.
Find all solutions, real and imaginary, of the equation:
x
12
x
2
3
4
x
12
4
x 2  3
x
x
12
4
x2  x  x2  2  x2  3  x2 
x
x
x 3  12  3 x 2  4 x
x 3  3 x 2  4 x  12  0
x  3, 2i , or  2i
x x  3   4x  3   0
2
x
2

 4 x  3   0
x2  4  0
2
x  4
or
x 3 0
x 3
x  4
x  2i
Therefore the solution is x  3, 2i , or  2i
14.
(a) 1
Given the function f ( x )  4x 2  3 find
(a) f(1)
f 1  412  3
 43
(b)
1
(b) f(0)
-3
(c) 1
f 0  402  3
 03
 3
(c) f(-1)
f  1  4 12  3
 43
1
15.
Find the domain and range for the function graphed below
.
5 y
The domain is all real numbers.
The range is all real numbers
greater than -3 i.e. y ≥ -3
–5
5
x
Range is y ≥ -3
–5
Math 098 Test 4 Review Solutions
Domain is all real numbers
Page 4
16.
Which of the following is a function?
A. {(-5,-3), (-3,-5), (5,5)} yes
B. {-5, -3, 1, 5} no, there are not x and y sets
C. {(-5,-3), (1,5), (1,-5), (5,1)}
D. {(-5,-3), (-3,1), (-5,5)}
17.
no – there are two solutions for x = 1
no – there are two solutions for x = -5
If a rocket is launched vertically upward from ground level with an initial velocity of
96 feet per second, then its height h after t seconds is given by the equation
h  16t 2  96t . What is the maximum height reached by the rocket?
The maximum height will occur at the vertex, and it will be the value h at that point.
b
96
, we get x  
Vertex: At x  
or x = 3
2a
 32
When x = 3,
h  1632  963
 144
The maximum height is 144 ft.
18.
Graph the parabola: y   x 2  6
Compare this parabola’s equation to the general form of the
parabola y  ax 2  bx  c
It has the typical shape of the graph y  x 2
This parabola is concave down because the coefficient
of the x2 is negative.
It is shifted vertically (UP) by 6 units (by the “+6”)
y-intercepts: when x = 0, y = 6, so it has a y intercept at
(0, 6)
x-intercepts: when y = 0





 x2  6  0
x2  6

x 6

So the x intercepts are 0, 6 and
19.
A
Graph:
0,  6 
y  x  12  2
Compare this parabola’s equation to the standard form of
the parabola y  ax  h 2  k




It has the typical shape of the graph y  x 2
This parabola is concave up because a is positive
The vertex of this parabola is h, k   1, 2
y-intercepts: when x = 0,

3
so it has a y intercept at (0, 3)
When y = 0
y   12  2
x  12  2  0
x  12  2 which has no real solution
So there are no x intercepts.
Math 098 Test 4 Review Solutions
Page 5
144 ft
20.
Graph:
y  2x 2  8 x  1
Compare this parabola’s equation to the general form of the parabola
y  ax 2  bx  c
 This parabola is concave up because the coefficient of the x 2 is positive, and
steeper than y  x 2 but steeper, because the a = 2
x   84

Vertex: when x   2ba ,

When x  2 , y  2 2  8 2  1
 7
The vertex is  2,  7
y-intercepts: when x = 0, y = 1, so it has a y intercept at (0, 1)

x-intercepts: when y = 0, 2 x 2  8 x  1  0
 2
2
Using the Quadratic Formula x 
x
 b  b2  4ac
with a  2, b  8, and c  1:
2a
 8  82  4  2  1
22

 8  56
4

 8  2 14
4

2  4  14
4

 4  14
2

x  2 
14
2
56  4  14
 2 14

 3.9
or
2
14
2
 0.1
(-2, -7)
So the x intercepts are  0,  2 

Math 098 Test 4 Review Solutions
14
2
 and


Page 6
 0,  2 


14
2


