```Practice Problems, Test 3 M102
(1) Simplify:
(b) 3 125
81
(a)
(c)
4x 2
(d)
3
8x3
(2) Find the distance between the points (-2,4) and (4,-4).
(3) Find the missing side of the right triangle:
15
9
b
(4) Rewrite the expression without rational exponents, and simplify, if possible:
(a) 4
1
2
(b) (8)
1
3
3
2

(c) 25
(a)
50
(b)
28
(c)
3
40 (d)
3
54
(e) 5 2  18
(a)
2( 2  3 3)
(b) (2 5  3 2)(3 5  4 2)
(c) Multiply 2  3 by it's conjugate.
(7) Rationalize the denominator:
(a)
3
2
(b)
2 3
3 5
(a) r  25r  144
(b)
z 1  z  2
(d) x
4
3
4
(e)  
9

5
2
1
1
(f) x 2 x 4
(9) Simplify the square root using complex numbers:
(a)
16
(b)
13
(10) Perform the complex operations:
(a) (3  4i)  (2  5i)
(b) 2i  (4  3i)
(c) (2  5i )(3  6i )
(d) multiply 4 - 3i by it's conjugate.
(e) divide:
2  5i
2  3i
Solutions:
81 = 9
(since 92 = 81 )
3
(since 53 = 125)
(1) (a)
(b)
4 x 2  (2 x) 2  2 x.
(c)
(d)
125 = 5
3
8 x3  3 (2 x)3  2 x.
(2) For this and all such problems, you need to know the distance formula between two
points d =
( x1  x0 )2  ( y1  y0 )2 .
Here, the points are (-2,4) and (4,-4), so the distance between them is:
(4  2) 2  (4  4) 2  62  82  100  10.
(3) We use the Pythagorean Theorem to find b
a 2  b2  c2 ,
c  a 2  b2 ,
b  c2  a2 .
a 2  b2  c2 ,
c  a 2  b2 ,
b  c2  a2 .
b  152  92  225  81  144  12.
(4)
1
n
The crucial rules here are: x  n x ,
m
and: x n  ( n x )m (remember, for fractional exponents - the
numerator is the power, and the denominator is the root).
1
2
(a) 4  4  2.
1
(b) (8) 3  3 8  2.
(since (-2)3 = -8)
(c) Negative exponents send stuff on the top to the bottom and the bottom to the top:
m
m

1
1
n
x  m,
 xn.
m

xn x n

25
3
2

1
25
3
2

1
1
1


.
3
( 25)3 5 125
4
(d) x 3  3 x 4 (The 3 is in the denominator, so it the root, the 4 is in the numerator, and
so it is the power - note that with variables, we like to put the power inside of the radical).
(e) Here, the negative exponent 'flips' the fraction (since the top goes to bottom and
bottom to top - then we take both numbers to the positive version of the exponent:
4
 
9

5
2
5
2
5
2
9
( 9)5 35 243
9
   5 
 5
.
5
2
32
(
4)
4
42
(f) Multiplying same base, different exponents - we have to add the exponents.
1
1
1 1

4
x2 x4  x2
3
 x4.
(5) Here, we need to pull out some roots.
50  5 2
(a)
25 2
28  2 7
(b)
4 7
3
(c)
3
40  2 3 5
83 5
3
(d)
3
54  3 3 2
27 3 2
(e) Here, we need to simplify one of the radicals (by pulling out roots) , then combine
them using the rule: a n x  b n x  (a  b) n x
5 2  18  5 2  3 2  2 2.
(6) Radical operations are like polynomial operations - in these two problems, we need to
distribute of use FOIL. Note here that I also make use of the fact that a * a  a.
(a)
2( 2  3 3)  4  3 6  2  3 6.
(b)
F
O
I
L
(2 5  3 2)(3 5  4 2)  6(5)  8 10  6 10  12(2)  6  2 10.
(c) The conjugate of 2  3 is 2  3 , and we need to use the difference of squares
formula (a + b)(a - b) = a2 - b2.
2
(2  3)(2  3  22  3  4  3  1.
(7) Rationalizing the denominator means eliminating the radical in the denominator - the
techniques for doing so are different in these two problems - in the first, we just multiply
by the square root that appears in the denominator. In the second, we have to multiply by
the conjugate of the denominator.
(a)
3
3

2
2
2 3 2

2
2
(b) Here is where we have to use the conjugate of the denominator.
2  3 3  5 6  2 5  3 3  15 6  2 5  3 3  15 6  2 5  3 3  15



.
95
4
3 5 3 5
32  ( 5) 2
(8) (a)
r  25r  144,
( square)
r 2  ( 25r  144) 2 ,
r 2  25r  144,
 25r  144
(quadratic : move everything to one side).
r 2  25r  144  0, ( factor :)
(r  16)(r  9)  0 ( set factors  0).
r  16  0, r  9  0,
 36  36  9  9
r  16, r  9.
9  25(9)  144 ?
9  225  144,
9  81.
16  25(16)  144 ?
16  400  144,
16  256.
So both candidate solutions work.
(b) This is a 'double radical' equation - what you have to do is isolate one radical and
square, then simplify, isolate the other radical and square, then solve for the variable. I'm
going to isolate z  1 first:
z 1  z  2
z 1  2  z

z 1

2
( square)
 (2  z )(2  z ),
z  1  4  2 z  2 z  z , ( Isolate
z  4  4
z
3  4 z
3
 z
4
2
3
  
4
9
 z.
16
Check :
z)
( square)
 z ,
2
9
9
1 
 2?
16
16
25
9

 1?
16
16
25
9

 1?
16
16
5 3
  1.
4 4
Yes.
(9) The key idea here is to 'pull the i out' - the square root of -1 is i.
(a)
16  16i  4i.
(b)
13  13i
(10) The key idea with complex operations is to 'treat i as a variable', and thus these
operations come down to combining like terms, using distribution and FOIL, and the
special property i2 = -1.
(a) (3  4i)  (2  5i) = 5 + i
3+2 (-4+5)i
(combine like terms)
(b) 2i  (4  3i) = 2i - 4 + 3i = -4 + 5i.
(combine like terms)
(c)
F
O
I
L
(2  5i )(3  6i ) = 6 + 12i - 15i - 30i2
-30(-1)
30
= 36 - 3i.
(d) The rule for multiplying complex conjugates is slightly different than the regular
difference of squares formula, owing to the fact that i2 = -1 alters the simplification:
(a  bi)(a  bi )  a 2  b2
The conjugate of 4 - 3i is 4 + 3i, and
(4  3i)(4  3i)  42  32  25.
(e) In order to divide by a complex number, we have to multiply by the conjugate of the
denominator:
2  5i 2  3i 4  6i  10i  15i 2 4  16i  15(1) 11  16i
11 16
*



   i.
2
2
2  3i 2  3i
2 3
13
13
13 13
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