Download IAD Solution Manual 2015

Intermediate Algebra Diagnostic Test (IAD)
Selected Solutions to the Sample Test Questions found in the CSUS Mathematics and
Statistics Diagnostic Test Study Guide
Note:
à
This Solution Manual is designed to help you understand how the answers to the IAD
Sample Test in the above Study Guide were found.
à
This booklet is not designed to stand alone but to be used AFTER you have tried the
questions in the Sample Test.
à
There is often more than one way to get to an answer and no attempt has been made
to show all possible ways to a solution.
Topic I:
Elementary Operations
Scientific Notation is a way of writing a number so that only one non-zero digit is to the left of
the decimal point (that is, the one’s place), the rest of the digits are to the right of the decimal,
and the number is times a power of 10. Scientific notation is useful for very large or very
small numbers.
For example, 232000 is written as 2.32 x 105 (the decimal moved to the left so the
exponent, 5, is positive)
–4
0.00026 is written as 2.6 x 10
(the decimal moved to the right so
the exponent, 4, is negative)
Remember when multiplying numbers to add the exponents and when dividing to subtract.
The list of exponent rules is at the beginning of Topic III.
1.
Express in Scientific Notation:
(5.7 x 10 ) ( 2.3 x 10 )
6
−3
Separate the numbers from the powers of 10. Then multiply the numbers and then the
powers of 10 (using the exponent rules).
( 5.7 x 2.3) x 10
( )
6+ − 3
This yields 13.11 x 103 which is not in scientific notation. Move the decimal one place
1.311 x 104
to the left and add one to the exponent 3. The answer is:
2.
Express in Scientific Notation:
3.81 x 10 − 5
1.05 x 10 − 7
⎛ 3.81⎞ 10− 5
Again, separate the numbers from the powers of 10. ⎜
. Divide the
x
⎝ 1.05 ⎟⎠ 10−7
numbers and subtract the powers of 10 to get: 3.63 x 10
( ) = 3.63 x 102
− 5− −7
Notice we rounded the answer to two decimal places and that this answer is in
scientific notation.
1
3.
4 x2y 3 9
⋅
⋅
x 24 y 4
To simplify fractions look for factors common to both the numerator and the
denominator of the fractions.
Simplify:
4 ⋅ x ⋅ x⋅ y ⋅y ⋅y ⋅ 3 ⋅3
=
x ⋅ 4 ⋅ 3 ⋅2⋅ y ⋅ y ⋅ y ⋅ y
4.
3x
2y
(
)
(
)
2 ⎡⎣ y − x − 5 ⎤⎦ − 3 ⎡⎣ 4 − x − y ⎤⎦
Follow the order of operations carefully! Remember PEMDAS?
Add and simplify:
(
)
(
)
2 ⎡⎣ y − x − 5 ⎤⎦ − 3 ⎡⎣ 4 − x − y ⎤⎦ =
2 ⎡⎣ y − x − 5 ⎤⎦ − 3 ⎡⎣ 4 − x + y ⎤⎦ =
2y − 2x − 10 − 12 + 3x − 3y =
( 2y − 3y ) + ( −2x + 3x ) + ( −10 − 12) =
−y + x − 22
then rearrange to get x − y − 22
To evaluate expressions for specific values of the variables, substitute the given number for
each variable in the expression. Hint: put any negative values in parentheses before
substituting so that you don’t mess up the signs.
5.
( )
If x = ( – 1) and a = 2, then
x3
−1
⋅ a4
a2 x 2
=
Before you substitute here you can simplify the expression. Remember that
1
1
m−1 =
so x −3 = 3 . Now the fraction looks like this and can be simplified to
m
x
a4
x 3a 2 x 2
=
a2
x5
Now substitute in the values for x and a.
22
( −1)
5
=
4
= −4
−1
2
6.
If a = – 3, evaluate 1− a − 2 a + 2 + 1
Absolute Value represents the distance from zero, so it is always positive.
( )
( )
1− −3 − 2 −3 + 2 + 1 =
1+ 3 − 2 −1 + 1 =
()
4 −2 1 + 1= 3
Topic II:
Rational Expressions
To add or subtract rational expressions (fractions) you must have common denominators
first. Then you can add and subtract the numerators.
7.
Add and simplify:
First write 2x as
xy 2 − x 2 y
xy + y 2
+ 2x
2x
and factor the first expression:
1
( ) + 2x
1
y (x + y)
xy y − x
Now simplify by canceling the y from numerator and denominator. Then make
common denominators and add the numerators.
( ) + 2x ⋅ ( x + y ) =
1 (x + y)
y (x + y)
xy y−x
xy − x 2 + 2x 2 + 2xy
=
x+y
3xy + x 2
x+y
You could factor out an x from the numerator but it will not cancel with anything in the
denominator (because of the + )
8.
Add and simplify:
xy 3
y 5 + xy 4
+
y−x
xy + y
xy 3
y 5 + xy 4
+
=
y−x
xy + y
(
(
)
y4 y + x
xy 3
+
=
y−x
y x +1
)
3
After the y cancels, the common denominator is
(
(
) (
)
)
(
) (
) (
)
)
( y − x ) ( x + 1)
x + 1 y3 y + x y − x
xy 3
⋅
+
⋅
=
y−x
x +1
x +1
y−x
(
x 2 y 3 + xy 3 + y 5 − x 2 y 3
=
y − x x +1
(
(
9.
)(
)
y 5 + xy 3
y − x x +1
)(
)
Simplify:
y−x
2 + 2x 2
⋅
4x 2
x2 − y 2
Since we are multiplying (and not adding) rational expressions, we do not need
common denominators!
y−x
⋅
2 + 2x 2
4x 2
x2 − y 2
=
( y − x ) 4x
2 (1+ x ) ( x + y ) ( x − y )
Factor everything you can first.
2
2
Notice the two factors:
( y − x ) in the numerator and ( x − y ) in the denominator.
These are called opposites (the same way that 2 and – 2 are opposites. You can
factor a (– 1) from y − x . Then it becomes y − x = −1 x − y . Now we can
(
)
(
) ( )(
)
cancel…
( −1) ( x − y ) 4 2x
−2x
=
2 (1+ x ) ( x + y ) ( x − y ) (1+ x ) ( x + y )
2
2
2
2
10.
Simplify:
1
x+3
3x − 1
− 2
+ 2
x+2
x + 3x + 2
x −1
First factor the denominators to find the LCD (Least Common Denominator).
1
x+3
3x − 1
−
+
x+2
x + 2 x +1
x +1 x −1
(
)(
) (
)(
)
(
)(
)(
)
The LCD is x + 2 x + 1 x − 1 (ie. the most
of any factor in one fraction). Now make equivalent fractions.
4
(
(
)(
)(
)
) (
(
) (
)
) (
(
) (
)
)
x +1 x −1
x −1
x+2
1
x+3
3x − 1
⋅
−
⋅
+
⋅
=
x + 2 x +1 x −1
x + 2 x +1
x −1
x +1 x −1
x+2
(
)(
)
x 2 − 1− x 2 + 2x − 3 + 3x 2 + 5x − 2
( x + 2) ( x + 1) ( x − 1)
)(
=
x 2 − 1− x 2 − 2x + 3 + 3x 2 + 5x − 2
=
x + 2 x +1 x −1
(
)(
)(
)
(
3x 2 + 3x
=
x + 2 x +1 x −1
(
3x
3x
or 2
x + x−2
x + 2 x −1
)( )
3x ( x + 1)
=
( x + 2) ( x + 1) ( x − 1)
11.
)(
)(
)
⎛ 3 1⎞
⎜⎝ y + 4 ⎟⎠
Simplify:
⎛
1⎞
⎜⎝ y − 4y ⎟⎠
First separate the numerator and denominator and find an LCD for each.
3 1
3 4
1 y
12 + y
⋅ + ⋅ =
The common denominator for + is 4y so,
.
y 4
y 4 4 y
4y
1
The common denominator for y −
is 4y so,
4y
⎛ 3 1⎞
2
⎜⎝ y + 4 ⎟⎠
2y + 1 2y − 1
y 4y
1
4y − 1
⋅
−
=
which factors to
. Now
1 4y
4y
4y
4y
⎛
1⎞
⎜⎝ y − 4y ⎟⎠
(
is now
(
)(
)
12 + y
4y
4y
which can be multiplied by
to yield
4y
2y + 1 2y − 1
)(
)
4y
(
12 + y
4y
12 + y
4y
⋅ 1 =
4y
2y + 1 2y − 1
2y + 1 2y − 1
1
4y
)(
)
(
)(
)
or
12 + y
4y 2 − 1
5
12.
−1 x
+
x +1 2
Simplify:
1
3
−
x +1 x
Separate the numerator and denominator and find the LCD of each. The numerator’s
LCD is 2(x + 1) and the denominator’s LCD is x(x = 1). Now make equivalent
x2 + x − 2
−1 ⎛ 2 ⎞ x ⎛ x + 1⎞
+
2 x +1
x + 1 ⎜⎝ 2 ⎟⎠ 2 ⎜⎝ x + 1⎟⎠
fractions and combine to get
=
−2x − 3
1 ⎛ x⎞
3 ⎛ x + 1⎞
− ⎜
⎜
⎟
⎟
x x +1
x + 1 ⎝ x⎠
x ⎝ x + 1⎠
(
)
(
)
Writing this horizontally and dividing fractions yields
x + 2 x −1
x + 2 x −1 x x +1
−2x − 3
÷
becomes
⋅
. This simplifies to
−2x − 3
2 x +1
x x +1
2 x +1
(
(
)(
(
)(
)
( )
( )
( x + 2) ( x − 1) ⋅ x ( x + 1) = − x ( x + 2) ( x − 1) .
− ( 2x + 3 )
2 ( 2x + 3 )
2 ( x + 1)
Topic III:
)
)
(
)
Exponents and Radicals
Recall some properties of exponents:
1
1
x −n = n
example: 2−3 = 3
x
2
n
4
⎛ a⎞
an
=
⎜⎝ b ⎟⎠
bn
⎛ 2⎞
24
=
⎜⎝ 3 ⎟⎠
34
( ab)
( 3x )
n
(a )
m
x
= anbn
a m a n = a(
am
a
n
n
=a
x=
(2 )
= amx
m+ n
3
)
x5
x
1
x n
Evaluate:
14.
Simplify:
5
= 32 x 2
= 215
x3 ⋅ x 4 = x7
m− n
13.
2
3
8
−2
3
10 ⋅ 20
2
= x3
5=5
1
3
8
−2
3
( )
= 23
−2
3
( 31 ⋅ −2 3 ) = 2−2 =
=2
1
2
2
=
1
4
10 ⋅ 20 = 10 ⋅10 ⋅ 2 = 10 2
6
15.
16x 2 + 24y 2 + 36z 2
Simplify:
(
)
( 4x
16x 2 + 24y 2 + 36z 2 = 4 4x 2 + 6y 2 + 9z 2 = 2
16.
(
1 ⎞
⎛
2
2
⎜⎝ 3ab ⎟⎠ 4a b
Simplify:
ab
3⋅ a
1
⋅b 2
⋅ 4−2 ⋅ a−4 ⋅ b−2
a ⋅b
3
3
)
−3
18.
Simplify for a > 0:
2
2
)
−2
=
3⋅b
( )
1 + −2
2
42 ⋅ a4 ⋅ b
(x y ) (x y )
Simplify:
+ 6y 2 + 9z 2
2
2
17.
2
−6
2
3
3
=
16a 4b
2
3
2
⋅b
=
3
2
3
16a 4b3
−6+12) ( 4+ − 4 )
x −6 y 4 x12 y −4 = x(
y
= x6 y 0 = x6
−2
3
8a3 + a 32a + 16a3
3
22 ⋅ 2 ⋅ a2 ⋅ a + a 42 ⋅ 2 ⋅ a + 23 ⋅ 2 ⋅ a3 = 2a 2a + 4a 2a + 2a 3 2 = 6a 2a + 2a 3 2
19.
Simplify:
⎛ 43 x 3 2 y 4 3 ⎞
⎜
⎟
⎜⎝ 16x −12 y 2 ⎟⎠
⎛ 4 2 ⋅ 4 ⋅ x 3 2 ⋅ x 12 ⎞
⎜
⎟
⎜⎝ 4 2 ⋅ y 2 ⋅ y −4 3 ⎟⎠
−1
2
−1
2
⎛
(3 + 1 ) ⎞
4⋅ x 2 2 ⎟
⎜
=
⎜ ⎛⎜⎝ 21 + − 4 3 ⎞⎟⎠ ⎟
⎝ y
⎠
−1
2
=
4
−1
2
(x )
2
⎛ y 23 ⎞
⎝
⎠
−1
2
1
y 3
=
2x
−1
2
Simplify:
a3 x + 2
a 4− x
a3 x + 2
3 x + 2− ( 4− x )
=a
= a3 x + 2− 4+ x = a 4 x − 2
4− x
a
21.
Simplify:
z 2 x −4 y 3
y −2 zx −2
z2 x2y 3 y 2 z ⋅ z ⋅ x2 ⋅ y 5 y 5z
=
= 2
x4z
x
x2 ⋅ x2 ⋅ z
22.
Express without radical signs:
20.
23.
Express without radical signs:
24.
Solve for y:
2x =
8x = 5 − 3y
25.
Simplify:
5 − 3y
4
x2 ⋅ 5 x
3
z
x
7
2
3
z
7
⋅x
3
1
5
=x
( 2 3 + 15 ) = x(1015 + 315) = x1315
7
= ⎛z 3⎞
⎝
⎠
1
2
=z
7
6
Multiply both sides by 4 to clear the fraction.
8x − 5 = −3y
10 x
2x
3
y=
8x − 5
−3
y=
5 − 8x
3
x
10 x ⎛ 10 ⎞
= ⎜ ⎟ = 5x
x
2
⎝ 2⎠
7
26.
Rationalize the denominator:
denominator by the conjugate:
2
3− 7
Topic IV:
⋅
3+ 7
3+ 7
=
2
(
3+
3−7
2
3− 7
Multiply the numerator and
( 3 + 7)
7) 2 ( 3 + 7) −
=
=
−2 ⋅ 2
3− 7
2
Linear Equations and Inequalities
There are 2 methods of solving a system of linear equations (ie. 2 equations with 2
variables). Either method can be used, so chooses the one that seems easier for the
problem. You will get the same answer with either method.
Substitution: solve one equation for one of the variables and substitute that expression into
the other equation. It is a good idea to put parentheses ( ) around the expression being
substituted. This method is best when one equation can be solved for one variable without
getting fractions.
Elimination: multiply each equation by a number so that when you add the equations
together, one variable is eliminated. This method is best when solving for any variable
creates fractions.
REMEMBER: You must solve for both variables. So after you find one answer, solve for the
other (by either method).
27.
Solve:
⎧1
⎪ x + 3y = −5
⎨2
⎪5x − 2y = 14
⎩
First, multiply the first equation by 2 to clear the
⎧ x + 6y = −10
You can now use either
⎨
⎩5x − 2y = 14
method to solve it. Using the Elimination method, multiply the first equation by ( – 5).
−5x − 30y = 50
fractions. Now the problem looks like this:
5x − 2y = 14
−32y = 64
Now we can use substitution to find x.
y = −2
( )
x + 6 −2 = −10
x − 12 = −10
x=2
28.
So, the answer is x = 2, y = – 2 or the point:
⎧ 4x = 5y
First, rewrite so it looks like
⎨
⎩ 2x + 3y = 22
Now, use Elimination again. Multiply the second equation by
Solve:
( 2,− 2)
⎧ 4x − 5y = 0
⎨
⎩ 2x + 3y = 22
– 2.
8
4x − 5y = 0
−4x − 6y = −44
Substitute y = 4 into the first equation to get…
−11y = −44
( )
4x = 5 4
y=4
x=5
So the solution is x = 5, y = 4 or the point:
( 5,4)
For #29, remember that when solving problems with square roots, be careful to square both
sides of the equation and always plug your answers back into the original equation to check
for extraneous solutions. Some answers will not work!!
29.
Solve for x:
2 − 1− x = −3
You could square both sides now but it is messier. It’s better to add – 2 to both sides
first.
2 − 1− x = −3
− 1− x = −5
30.
1− x = 5
1− x = 25
x = −24
Then check the answer.
Solve for m:
17 − 19m > −4 m + 5
(
)
Remember that when multiplying or dividing by a negative number, the inequality sign
changes direction. (Adding or subtraction negative does not change the direction of
the inequality sign)
17 − 19m > −4m − 20
−15m > −37
37
m<
15
31.
(
) (
)
(
3 5 − 4x − 3 − 7x = − 2 + 9x
Solve for x:
)
15 − 12x − 3 + 7x = −2 − 9x
12 − 5x = −2 − 9x
4x = −14
−14
−7
x=
x=
4
2
Topic V: Quadratic Polynomials, Equations, Inequalities
32.
Multiply:
(x
2
)(
− 5x 4x 2 − 3x + 2
)
You can use the distributive property or rectangle multiplication.
9
4x2
2
4x4
– 3x3
+2x2
– 5x
– 20x3
+ 15x2
– 10x
4x 4 − 23x 3 + 17x 2 − 10x
Factor completely:
4a 2 − 9a 2b 4
First look for factors in common for each term. Here it is a 2 .
a 2 4 − 9b 4 This is a difference of squares and yields: a 2 2 − 3b 2 2 + 3b 2
(
34.
+2
x
Now combine like terms to get:
33.
–3x
)
(
x2 +
Solve for x:
5
3
x=
2
2
)(
)
First multiply both sides by 2 to clear the fractions.
2x 2 + 5x − 3 = 0
( 2x − 1) ( x + 3) = 0
2x − 1= 0
1
x=
2
35.
or
x+3=0
or
x = −3
x4 + x2 − 2 = 0
Solve for x:
Let m = x 2 . Then the question becomes how to factor m2 + m − 2 = 0 .
This is much easier. m + 2 m − 1 = 0 which gives us m = – 2 or m = 1.
(
)(
)
2
Now substitute x for m and solve for x.
36.
x 2 = −2
or
x = ± i 2 or
x2 = 1
x = ±1
If you solve this equation by completing the square, what do you add to each side of
x 2 + 3x = 4
the equation?
To complete the square means to take the x2-term and the x-term and form a perfect
square by adding something. This “something” can be calculated by dividing the
x-term coefficient by 2 and squaring the result.
2
⎛ 3⎞
9
The coefficient of the x-term is 3. So, divide 3 by 2 and square it. ⎜ ⎟ =
4
⎝ 2⎠
So the answer to this question is to add
9
to complete the square. This is how it
4
would solve from there.
10
x 2 + 3x +
9
9
= 4+
4
4
2
⎛
3⎞
25
⎜⎝ x + 2 ⎟⎠ = 4
x+
37.
−3 5
+
2 2
x=1
3
5
=±
2
2
x=
or
or
−3 5
−
2 2
x= −4
x=
2x 2 + 4x + 1= 0
Solve for x:
Solving this quadratic equation requires the Quadratic Formula. Any equation in the
form ax 2 + bx + c = 0 can be solved by this formula:
x=
−b ± b 2 − 4ac
.
2a
In this problem a = 2, b = 4, and c = 1.
x=
()
2 2
x=
− 4 ± 16 − 8
4
x=
−4± 8
=
4
x=
−4±2 2
4
x=
x=
( )( )
− 4 ± 42 − 4 2 1
(
2 −2± 2
)
4
−2± 2
2
38.
39.
Omitted
Omitted
40.
Simplify:
i
2 − 5i
Recall that i = −1 . This gives us i 2 = −1, i 3 = − i, and i 4 = 1. So any power of i can
be simplified to i, – 1, – i, or 1 by dividing the exponent by 4 and looking at the
remainder. For example, i 345 = i 86 ⋅ 4 ⋅ i 1 = 1⋅ i = i .
i
To simplify
we must multiply by the conjugate of the denominator (same ends,
2 − 5i
different sign between). Here the conjugate is 2 + 5i.
i
2 + 5i
2i + 5i 2
2i + 5i 2
2i − 5 − 5 + 2i
2
⋅
=
=
=
but
i
=
–
1,
so
2
2
2 − 5i 2 + 5i
4 + 25
29
4 − 25i
4 − 25i
11
41.
z=
( ) ( −4 ) − 4 ( 3) ( 2)
2 ( 3)
− −4 ±
2
z=
4 ± 16 − 24
6
z=
4± −8
4 ± 2i 2
which simplifies to
6
6
z=
42.
3z 2 − 4z + 2 = 0 Use the Quadratic Formula.
Solve for z:
(
2 2±i 2
)
The answer is: z =
63
2±i 2
3
1
1
1
=
Note that a cannot be 1 or − .
1+ 2a a − 1
2
The LCD of the denominators is (1 + 2a)(a – 1) so multiply each term by that to clear
the fractions.
3−
Solve for a:
(
)(
) ( ) (
− a − 1) − ( a − 1) = 1+ 2a
3 1+ 2a a − 1 − 1 a − 1 = 1 1+ 2a
(
3 2a 2
)
6a 2 − 3a − 3 − a + 1= 1+ 2a
6a 2 − 6a − 3 = 0
(
)=0
3 2a 2 − 2a − 1
3
2a − 2a − 1= 0
3
2
a=
a=
a=
43.
Now use the Quadratic Formula to solve for a
( ) ( −2) − 4 ( 2) ( −1)
2 ( 2)
2
− −2 ±
2± 4+8
4
(
2 1± 3
)
4
which simplifies to
So, a =
2±2 3
4
1± 3
2
(
)( )
First solve for m when ( 2m − 1) ( m − 5 ) = 0 .
For what values of m is 2m − 1 m − 5 > 0 ?
1
, m=5
2
Now find these numbers on a number line. These are called critical points. We need
to see what happens on each side of these critical points. Choose test points 0, 1, and
6.
½
0
m=
5
1
6
12
Substitute the test numbers into the inequality to see whether the product is greater
than 0 or less than 0.
2 ⋅ 0 − 1 0 − 5 = 5 which is greater than 0 5 > 0
(
)( )
( 2 ⋅1− 1) (1− 5) = − 4 which is less than 0
( 2 ⋅ 6 − 1) ( 6 − 5) = 11 which is greater than 0
−4<0
11 > 0
So, every number greater than 5 and every number less than ½ will yield a product
1
or m > 5
that is greater than 0. Therefore, the answer is m <
2
Topic VI:
44.
Graphing and Geometry
Find the equation of the line which passes through the points (1, 3) and (7, – 2).
3 − −2
y − y1
First we need to find the slope which is: m = 2
. So, in this case, m =
1− 7
x2 − x1
( )
−5
. Note: you can make either point x1, y 1 , but be careful
6
not to mix them up. Now choose a point (either one will work) and substitute it and the
slope into the formula: y = mx + b . Using the point (1, 3), we get
(
which simplifies to m =
⎛ −5 ⎞
3=⎜ ⎟ 1 +b
⎝ 6⎠
()
3+
5
=b
6
b=
23
6
)
So, the equation of the line is
−5
23
x+
. If you need the standard form of the line, multiply through by 6 and
6
6
move the x term to get: 5x + 6y = 23
y=
45.
Find the equation of the line with a slope of 3 which passes through the point (0, – 2).
b = −2
Using the information given and y = mx + b , we get −2 = 3 0 + b
()
So the equation of the line is y = 3x − 2 .
46.
Find the distance between points A and B on the graph.
First find the coordinates of the points on the graph. A = ( – 3, 1) and B = (2, 2).
Now use the distance formula to find the length of the line segment between the
points: d =
d=
47.
(x
2
− x1
) + (y
2
( 2 − ( −3)) + ( 2 − 1)
2
2
2
− y1
)
2
In this case,
d = 5 2 + 12
d = 26
What is the midpoint between the points ( – 5, 3) and (7, – 2)?
The midpoint is the average of the x’s and the average of the y’s.
⎛ x1 + x2 y 1 + y 2 ⎞ ⎛ −5 + 7 3 + −2 ⎞
⎛ 1⎞
=⎜
,
⎟ = ⎜ 1, ⎟
⎜ 2 ,
⎟
2 ⎠ ⎝ 2
2 ⎠
⎝ 2⎠
⎝
( )
13
48.
Find the quadratic function whose graph is shown in the figure.
A quadratic function graphs as a parabola and has the form y = ax 2 + bx + c . If the a
is positive the graph opens upward. If the a is negative, the parabola opens
downward. In this case the a is negative. When x = 0, then y = 2, so the point (0, 2)
can be used to find the c. So, our function now looks like: y = −ax 2 + bx + 2 . Using
both points (2, 0) and ( – 2, 0), we get a system of equations:
⎧0 = −a 22 + b 2 + 2
⎧0 = −4a + 2b + 2
⎪
or ⎨
⎨
2
⎩0 = −4a − 2b + 2
⎪0 = −a −2 + b −2 + 2
⎩
Adding these two equations together, we get:
0 = −8a + 4
( ) ()
(( ) )
1
=a
2
49.
( )
1
for a and solving for b we find that b = 0
2
1
So, our function is now y = − x 2 + 2
2
Find the slope of the line, l, in the figure given.
Locate two points on the line. The easiest to find are (0, – 3) and (2, 0). Using the
0 − −3
y − y1
3
slope formula: m = 2
, we get m =
= , which is the slope of the line, l.
2−0
2
x2 − x1
Substituting
( )
50.
Graph the line y = 4
This says for any x the y value is 4. This is a horizontal line, whose graph is in the
answer key on page 27.
51.
Graph the function:
3x – 4y – 5 = 0.
⎛ −5 ⎞
3
5
x − . Find the point ⎜ 0,
on
4
4
4 ⎟⎠
⎝
the y – axis. Then move up 3 units and then 4 units to the right. Connect the two
points to make the line. The graph is in the answer key on page 27.
First solve the equation for y. This gives you y =
52.
Graph the equation: y = 2x 2 − 1
Note: this is NOT a straight line because the x is squared. Make a table with some
positive and negative values for x and calculate the y values. Then plot the points you
find.
The graph is in the answer key on page 27.
x
0
1
–1
2
–2
y = 2x 2 − 1
–1
1
1
7
7
14
53.
54.
OMIT
Find x.
This is a right triangle so you can use the Pythagorean Theorem to find x. The
Pythagorean Theorem is a 2 + b 2 = c 2 when a and b are the legs of a right triangle
and c is the hypotenuse (longest side). In this case, 7 2 + x 2 = 10 2 . Solving, we get
49 + x 2 = 100
x 2 = 51
x = 51
55 through 60. OMIT
Topic VII:
61.
Logarithms and Functions
( )
(
)
If f x = 3x 2 − 4x + 5, then f u − 2 =
(
)
(
)
2
(
)
f u−2 =3 u−2 −4 u−2 +5
(
) (
)
= 3 u 2 − 4u + 4 − 4 u − 2 + 5
Substitute (u – 2) for each x in the function.
= 3u 2 − 12u + 12 − 4u + 8 + 5
= 3u 2 − 16u + 25
62.
()
If f h = h +
3
+ 2 find f −3
h
( )
( )
3
+2
−3
= −3 − 1+ 2
= −2
f −3 = −3 +
Substitute (– 3) for each x in the function.
63.
Write logb a = x in exponential notation.
This is the definition of a logarithm. logb a = x
64.
65.
⇔
a = bx
Solve for x: 43 x = 16
First write both sides in the same base. Here you can just change the 16 to 42.
2
43 x = 4 2 . This is only true when 3x = 2. That is, when x = .
3
Solve for x: log x 125 = 3
First take this out of log notation and put it into exponential notation: 125 = x 3 .
125 = 53 so 53 = x 3 You can see that x must equal 5.
66.
Evaluate:
log64 16 = x
log64 16
16 = 64 x
First, set this equal to x then take it out of log notation.
( )
4 2 = 43
x
2 = 3x so, x =
2
3
15
67.
( 2x + 5) = 5
68.
(
)
log5 2x + 5 = 3
Solve for x:
Take this out of log notation.
3
2x + 5 = 125
2x = 120
x = 60
Solve for x:
log2 x + log2 x − 7 = 3
(
)
( )
Remember the property of logs that says: log a + log b = log ab
(
So,
)
log2 ⎡⎣ x x − 7 ⎤⎦ = 3
x 2 − 7x = 23
x 2 − 7x − 8 = 0
( x − 8) ( x + 1) = 0
x = 8 or x = −1
But x = – 1 does not work since you cannot take the log of a negative number. so the
answer is x = 8 is the only answer.
69.
⎛ 1⎞
log5 25 − log4 ⎜ ⎟ − log7 7
⎝ 16 ⎠
We cannot combine these by the properties of logs because the bases are different.
So we must evaluate each separately. Set each term to equal a different variable.
⎛ 1⎞
log5 25 = x
log4 ⎜ ⎟ = y
log7 7 = z
⎝ 16 ⎠
Simplify:
1
= 4y
16
2
x
5 =5
4 −2 = 4 y
2= x
−2 = y
So, x – y – z = 2 – ( – 2 ) – 1
⎛ 1⎞
log5 25 − log4 ⎜ ⎟ − log7 7 = 3
⎝ 16 ⎠
25 = 5 x
Topic VIII:
70.
7 = 7z
71 = 7 z
1= z
Word Problems
You can exchange $2.00 for £1.7. How many pounds (£) can you get for $15?
1.7 x
=
Use ratios:
Then cross multiply and solve for x.
2 15
25.5
2x = 1.7 15
x=
x = £12.75
2
( )
71.
The side of a square is doubled. How much does the area change?
If the square has a length of x, then A = x 2 . When we double x, it becomes 2x.
16
So, now the area is:
( )
A = 2x
2
A = 4x 2
So the area is four times larger.
72.
If 73% of a number is 365, what is the number?
part
percent
=
For percentages like this use this ratio:
remembering that what
whole
100
comes after the word “of” is the whole.
365
73
=
So, this problem becomes
Then cross multiply.
whole 100
36500
73 ⋅ whole = 36500
whole =
The whole number is 500.
73
73.
10 is to 28 as 35 is to what number?
10 35
=
10x = 28 ⋅ 35
Use ratios again.
28
x
74.
28 ⋅ 35
10
x = 98
The product of two numbers is – 48. Their sum is 13. What are the numbers?
Product is the answer to a multiplying problem. Sum is the answer to an adding
problem. (Difference is the answer to a subtracting problem. Quotient is the answer
to a dividing problem)
Let x be one number and y be the other number.
x ⋅ y = −48 and x + y = 13
Solve this system of equations by substitution. y = (13 – x )
x 13 − x = −48
(
)
13x − x 2 = −48
0 = x 2 − 13x − 48
(
)(
0 = x − 16 x + 3
75.
x=
If x = 16, then y = – 3, and if x = – 3, then y equals 16.
)
A rectangle has an area of 40 and a perimeter of 26. What are the dimensions of the
rectangle?
Let L be the length of the rectangle and W be the width.
Area = LW and 2L + 2W = Perimeter
40 = LW and 2L + 2W = 26 Solve for W and use substitution to solve this system
of equations.
40 = L 13 − L
(
(
W = 13 − L
)
)
40 = 13L − L
2
So
L2 − 13L + 40 = 0
( L − 8) ( L − 5) = 0
L= 8 or L =5 If L = 8, then W = 5 and if L = 5, then W = 8
So the dimensions of the rectangle are 5 by 8.
76.
Five times a positive number is equal to the difference of the square of that number
and 36. What is the number?
17
Let n be the number. Then five times the number is 5n.
The square of the number is n 2 . The equation is then…
5n = n 2 − 36
0 = n 2 − 5n − 36
(
)(
0= n−9 n+4
)
n = 9 or n = −4
But n is a positive number so the answer is only 9.
I have tried to be clear and concise in these solutions. There is
often more than one way to solve a problem, so you can get the
right answer by a different method.
Please contact me with comments or suggestions on ways to
improve this solution manual.
My email address is: [email protected]
Thank you!
β
18