MATH 131:100 Quiz 1 Solutions
Since a solid foundation in algebra is very important for success in this course, I decided
to give pretty thorough explanations to these problems, at the risk of sometimes being too
pedantic, so just bear with me. Also, please note how I write mathematics. It is still
important for the reader to understand what you are saying. It is all too common these days
(and I was guilty of it too as an undergrad student believe me) for people to turn in a mess
of scribbles on a page with some things boxed as answers and arrows leading you around the
page to see the flow of the work, but you wouldn’t do something similar in an English class,
so it stands to reason that you shouldn’t do it in a Math class either. In whatever you do
in life, writing to be understood by your audience is extremely important.
Diagnostic Test: Algebra
1) Evaluate each expression without using a calculator
NOTE: Just to recall for you, some basic facts about exponents (you can also consult
Section 1.5, specifically pages 53-54). We will also cover these on the second day of class.
(i) If n is some integer (that is, n = 1, 2, 3, 4, . . . ) and a is some number, then
an = a · a · a · · · · · a,
where there are n number of terms in the product. For example, a4 = a · a · a · a.
(ii) We also define
1
a−n = n .
a
(iii) If x and y are numbers, then
ax+y = ax ay .
(iv) From the previous property, we can also see that
ax−y =
ax
.
ay
Why? Put the last two properties together as follows:
ax−y = ax a−y =
ax
.
ay
(v) Also,
y
x
(ax ) = axy = (ay ) .
1
(vi) For an √
integer n, we define the fractional power a n =
1
example 4 2 = 4 = 2.
(vii) From the last property, if p and q are integers, then
√
1
ap/q = (ap ) q = q ap .
√
n
a to be the n-th root. For
But we can also do this in reverse order (because of (v)):
p
√ p
ap/q = a1/q = q a .
(viii) Finally, if a and b are numbers and n is an integer, then (ab)n = an bn .
SOLUTION:
a) By (i),
(−3)4 = (−3)(−3)(−3)(−3) = 9 · 9 = 81.
b) Parentheses make a huge difference:
−34 = − [3 · 3 · 3 · 3] = − [9 · 9] = −81.
1
c) By (ii),
3−4 =
1
1
=
.
4
3
81
d) By (iv),
523
= 523−21 = 52 .
521
e) First by (ii), then by (viii),
−2 2
3
32
9
2
=
= 2 = .
3
2
2
4
Remember that
1
2
3
=
3
.
2
(You flip the numbers)
2) Simplify each expression. Write your answer without negative exponents.
a) The idea here is that we need to find a common factor to pull out:
√
200 −
√
32 =
√
25 · 8 −
√
(viii)
4·8 =
√
√
√ √
√
√
25 8 − 4 8 = 5 8 − 2 8
√
√ (viii) √
√ √
√
√
= (5 − 2) 8 = 3 8 = 3 4 · 2 = 3 4 2 = 3 · 2 2 = 6 2.
2
b) First notice by (v), (4ab2 )2 = 42 a2 b2 = 16a2 b4 . Then we group all of the factors
with the same base, and use (iii):
factor
3a3 b3
4ab2
2
= 3a3 b3 16a2 b4 = (3)(16)a3 a2 b3 b4 = 48a3+2 b3+4 = 48a5 b7 .
c) First we flip and then square the top and bottom (using (ii), (v), and (viii))
3x3/2 y 3
x2 y −1/2
−2
=
x2 y −1/2
3x3/2 y 3
2
=
x4 y −1
x
= 7.
3
6
9x y
9y
In the second expression from the right, the y −1 in the numerator comes from the fact that
2
y −1/2 = y (−1/2)(2) = y −1 , and similarly for the x3 term in the denominator. For the final
term, notice that x4 /x3 = x4−3 = x, and also that
y −1
1 1
1
=
= 7.
y6
y y6
y
3) Expand and simplify. FOIL? Hopefully you learned that in high school somewhere.
a) The first term is 3(x + 6) = 3x + (3)(6) = 3x + 18, and the second: 4(2x − 5) =
4(2x) − 4(5) = 8x − 20. So the whold expression equals
3x + 18 + 8x − 20 = (3 + 8)x − 2 = 11x − 2.
b)
(x + 3)(4x − 5) = x(4x) + (−5)x + 3(4x) + 3(−5) = 4x2 − 5x + 12x − 15 = 4x2 + 7x − 15.
c)
√ √
√
√
√ √
√
√
√
√
√
√ √
√
( a + b)( a − b) = a a + a(− b) + b a + b(− b) = a − ab + ab − b = a − b.
√
√
√
√
This kind of thing will be very important later: a − b is called the conjugate of a + b
(conjugate means you just reverse the sign in the middle).
2
d)
(2x+3)2 = (2x+3)(2x+3) = 2x(2x)+3(2x)+3(2x)+3(3) = 4x2 +6x+6x+9 = 4x2 +12x+9.
e)
(x + 2)2 = x2 + 2x + 2x + 4 = x2 + 4x + 4.
4) Factor each expression.
a) You need to remember the following formula:
a2 − b2 = (a + b)(a − b),
note this is very similar to (3c). Then since 4x2 = (2x)2 , and 25 = 52 , we have
4x2 − 25 = (2x + 5)(2x − 5).
b)
2x2 + 5x − 12 = (2x − 3)(x + 4).
Here, the only way to get 2x2 is by multiplying x and 2x so those should be your starting
terms. Then you need two numbers that multiply to give you 12, so either 1 and 12, 2 and
6, or 3 and 4. Then just play around with those and the plus and minus signs until you get
5x for the middle terms.
d) First factor out an x since you can.
x4 + 27x = x x3 + 27 .
Now there is another general formula that you should have in the back of your mind somewhere:
a3 + b3 = (a + b)(a2 − ab + b2 ).
From this (and the fact that 33 = 27), we see that x3 + 27 = (x + 3)(x2 − 3x + 9), so
x4 + 27x = x(x + 3)(x2 − 3x + 9).
e) First, factor out the x−1/2 as follows
3x3/2 −9x1/2 +6x−1/2 = x−1/2 3x3/2+1/2 − 9x1/2+1/2 + 6x−1/2+1/2 = x−1/2 (3x2 −9x+6).
We can factor out the 3 in the second term to get 3x−1/2 (x2 − 3x + 2), which is
3x−1/2 (x − 1)(x − 2).
f) Factor out the x and the y first to get x3 y − 4xy = xy(x2 − 4). Then as in (4a), we
have that
xy(x2 − 4) = xy(x + 2)(x − 2).
5) Simplify the rational expression.
In the first three at least, simply factor the numerator and denominator as much as you
can and cancel like terms.
a)
(x + 1)(x + 2)
x+2
x2 + 3x + 2
=
=
,
x2 − x − 2
(x − 2)(x + 1)
x−2
where we cancelled the x + 1 term in the numerator and denominator.
b)
2x2 − x − 1 x + 3
(2x + 1)(x − 1) x + 3
x−1
·
=
·
=
,
x2 − 9
2x + 1
(x + 3)(x − 3) 2x + 1
x−3
where we cancelled the 2x + 1 terms and the x + 3 terms in the last step.
3
c) First we factor the first term to get (x + 2)(x − 2) in the denominator, and then we
need to get a common denominator, which we get by multiplying the second term by x−2
x−2
(since this is just 1, we haven’t changed the value of the fraction).
x2
x+1
x2
x+1x−2
x2
(x + 1)(x − 2)
−
=
−
=
−
(x + 2)(x − 2) x + 2
(x + 2)(x − 2) x + 2 x − 2
(x + 2)(x − 2) (x + 2)(x − 2)
x2 − x2 − x − 2
x2 − x − 2
x+2
1
x2
−
=
=
=
.
=
(x + 2)(x − 2) (x + 2)(x − 2)
(x + 2)(x − 2)
(x + 2)(x − 2)
x−2
d) For problems like these, usually the easiest thing to do is to multiply numerator and
denominator by whatever will cancel the fractions in the denominator. So in this case, start
by multiplying top and bottom by xy to get
y
x
1
y
−
−
x
y
1
x
xy
=
xy
xy 2
x
xy
y
−
−
x2 y
y
xy
x
=
y 2 − x2
(y − x)(y + x)
−(x − y)(y + x)
=
=
= −(y+x) = −y−x.
x−y
x−y
y−x
To switch the sign in third term from the right, I just factored out a −1 (i.e. y−x = −(x−y)).
6) Rationalize the expression and simplify.
Remember the conjugate I told you about in Problem (3c)? Here
use it.
√ is where you√
a) Multiply numerator and denominator by the conjugate of 5 − 2 (which is 5 + 2):
√
√
√
√ √
√
√ √ √
√
√
√
10
5+2
10 5 + 2 10 √ √
√
√
= 10 5 + 2 10 = 2 5 5 + 2 10 = 5 2 + 2 10.
=
5−4
5−2 5+2
√ b) Here you multiply top and bottom by the conjugate of the numerator (which is
4 + h + 2):
√
√
4+h−2 4+h+2
4+h−4
h
1
√
√
√
=
=√
=
.
h
4+h+2
h 4+h+2
h 4+h+2
4+h+2
Pay attention to this problem; you will see these types of thing over and over again when
we talk about differentiation.
7) Rewrite by completing the square.
Here you want to add and subtract the same number so that you get something that
looks like (x + a)2 + b. In general, if you have x2 + cx + d, you will add and subtract the
2
2
2
number c4 because x + 2c = x2 + cx + c4 . So then you would have that
c2
c2
c 2
c2
2
2
x + cx + d = x + cx +
−
+d= x+
+ d−
.
4
4
2
4
And if you have a coefficient other than 1 in from of x2 , then simply factor it out first and
do the above process.
2
a) As suggested, add and subtract 14 = 41 , and you will get
1 1
x +x+1=x +x+ − +1=
4 4
2
2
1
x+
2
2
3
+ .
4
b) Factor out the 2 from the first two terms, but then you have to be a bit more careful
as we complete the square on the inside.
36 36
2(x2 − 6x) + 11 = 2 x2 − 6x +
−
+ 11 = 2 x2 − 6x + 9 − 9 + 11
4
4
= 2 x2 − 6x + 9 − 18 + 11 = 2(x − 3)2 − 7.
4
8)
a)
1
3
2
x + x = 14 − 5 =⇒ x = 9 =⇒ x = 9 = 6.
2
2
3
b) Cross multiply:
2x
2x − 1
=
⇒ 2x(x) = (2x − 1)(x + 1) ⇒ 2x2 = 2x2 + x − 1 ⇒ x = 1.
x+1
x
c) Factor:
x2 − x − 12 = (x − 4)(x + 3) = 0,
which implies that either x = −3 or x = 4.
d) You can’t factor this one, so use the quadratic formula. Recall that if you have
ax2 + bx + c = 0, then
√
−b ± b2 − 4ac
.
x=
2a
Consequently, taking a = 2, b = 4, and c = 1, we see that
√
√
√
−4 ± 2 2
−4 ± 16 − 8
2
=
= −1 ±
.
x=
4
4
2
f)
3|x − 4| = 10 ⇒ |x − 4| =
So either
x=4+
10
10
10
⇒x−4=
, or − (x − 4) =
.
3
3
3
10
22
=
,
3
3
or x = 4 −
10
2
= .
3
3
Solve each inequality. Write your answer using interval notation.
Recall that adding or subtracting the same number on all sides of an inequality keeps it
the same, and when multiplying or dividing by a negative number, you change the direction
of the inequality.
a)
−4 < 5 − 3x ≤ 17
−4 − 5 < −3x ≤ 17 − 5
−9 < −3x ≤ 12
−9
−3x
12
>
≥
−3
−3
−3
3 > x ≥ −4.
Or in interval notation [−4, 3) is the solution.
d) |x − 4| < 3 means that −3 < x − 4 < 3, or 1 < x < 7. Or (1, 7).
10) State whether each equation is true or false. If it is false, provide a counterexample.
NOTE: This is probably the most important problem of this entire exercise because most
of these are the common mistakes that students keep making. Remember these so that you
don’t fall into the same trap!
a) (p + q)2 = p2 + q 2 .
This is false. For example:
(1 + 2)2 = 32 = 9,
b)
√
ab =
but
12 + 22 = 1 + 4 = 5.
√ √
a b. This is true by the exponential rules.
5
c)
√
a2 + b2 = a + b. We could just as well state this as
This is false. For example:
√
√
1 + 3 = 4 = 2,
√
but
1+
√
√
a+b=
3=1+
√
√
a+
√
b.
3 > 2.
A quick note: Functions such that f (a + b) = f (a) + f (b) are called linear functions for
a reason. The only power x such that (a + b)x = ax + bx is x = 1.
d)This is false: it should be
1
TC
1
1 + TC
=
+
=
+ T.
C
C
C
C
A term has to appear in all parts of the numerator and denominator to be cancelled.
e)
1
1
1
= − .
x−y
x y
This is false as I commented on in part (c). An example is
1
1
= ,
4−2
2
but
1 1
1
− =− .
4 2
4
f)
1/x
1
=
.
a/x − b/x
a−b
This is true. To see this, multiply the fraction on the left hand side of the equation by
This does not change the value because xx = 1.
x
x.
Note: In general, if you are unsure if something is true or not, try plugging in numbers to
see if it works. But you should typically avoid using 0 or 1 to test your equations, because
this can lead you to some wrong conclusions.
6
Diagnostic Test: Analytic Geometry
Recall that in general, given two points (x1 , y1 ) and (x2 , y2 ), we can represent the line
passing through the two points by the equation
y − y1 = m(x − x1 )
where
y2 − y1
x2 − x1
We could equally well replace y1 and x1 by y2 and x2 in the first equation. This equation is
called Point-Slope Form.
m=
1) Find an equation for the line that passes through the point (2, −5) and
a) has slope −3.
We just use the Point-Slope Form equations with slope m = −3:
y − (−5) = −3(x − 2)
Which when we rearrange, gives us
y = −3x + 6 − 5 = −3x + 1
b) is parallel to the x-axis.
A line parallel to the x-axis will always be of the form y = c where c is some number. In
this case, since our line needs to pass through the given point where the y-coordinate is −5,
we say
y = −5
b) is parallel to the y-axis.
A line parallel to the y-axis will always be of the form x = c where c is some number. In
this case, we use
x=2
d) is parallel to the line 2x − 4y = 3.
First, rearrange this formula into one of the form y = mx + b, the usual formula for a
line. Do this by adding 4y to each side and subtracting 3 from each side to get 4y = 2x − 3.
Now divide by 4, and the equation is y = x/2 − 3/4. This line has slope 1/2, so we need
the equation of a line with slope 1/2 that passes through the point (2, −5). Again, we use
Point-Slope Form:
1
y + 5 = (x − 2)
2
or
1
1
y = x−1−5= x−6
2
2
2) Find an equation for a circle that has center (−1, 4) and passes through the point
(3, −2).
A circle of radius r centered at the point (a, b) is given by the formula
(x − a)2 + (y − b)2 = r2
so we just translate this into our problem. The only issue is that we need to know that
radius. To do this, just find the distance from the center (−1, 4) to the point (3, −2) which
lies on the circle. Recall the distance between two points (x1 , y1 ) and (x2 , y2 ) is
p
(x2 − x1 )2 + (y2 − y1 )2
7
Then we have
r=
p
(3 − (−1))2 + (−2 − 4)2 =
p
√
√
42 + (−6)2 = 16 + 36 = 52
So r2 = 52, and the formula for our circle is
(x + 1)2 + (y − 4)2 = 52
4) Let A = (−7, 4) and B = (5, −12) be points in the plane.
a) Find the slope of the line that containst A and B.
This is just the difference of the y-coordinates divided by the difference of the x-coordinates
as I said before:
−16
4
−12 − 4
=
=−
m=
5 − (−7)
12
3
b) Find an equation of the line that passes through A and B. What are the intercepts?
Use Point-Slope Form to find the equation:
4
y − 4 = − (x + 7)
3
d) Find the length of the segment AB.
This is just the distance between two points that I used in number 2 of this Diagnostic
Test, so the length is
p
p
√
√
(5 − (−7))2 + (−12 − 4)2 = 122 + (−16)2 = 144 + 256 = 400 = 20
Diagnostic Test: Functions
1) The graph of a function is given in your book.
a) f (−1) = −2
b) f (2) is roughly 2.8
c) f (x) = 2 when x = −3 and when x = 1.
d) Estimate the values of x such that f (x) = 0.
Roughly at −2.5 and 0.4. (It is fine if these aren’t the exact numbers you chose).
e) State the domain and range of f .
D(f ) = [−3, 3]
R(f ) = [−2, 3]
2) If f (x) = x2 , evaluate the difference quotient
f (2+h)−f (2)
h
and simplify your answer.
I think probably the most common mistake made here will be what should the value
of f (2 + h) be? Remember, whatever is inside the parenthesis is what you plug into the
function. So f (2 + h) = (2 + h)2 = 4 + 4h + h2 . Then we procede from there:
f (2 + h) − f (2)
(2 + h)2 − 22
4 + 4h + h2 − 4
4h + h2
=
=
=
=4+h
h
h
h
h
where in the last term, I cancelled the h in the denominator since both terms had an h in
the numerator.
8