Solutions to Quiz 6A
www.math.ufl.edu/˜harringt
October 5, 2007
1. Reduce and simplify
(a)
(3+2
i
)
5
7
i
(1 +
i
2
) =
(3+2
i
)
5
7
i
(1

1) =
(3+2
i
)
5
7
i
*
0 = 0
(b)
i
21
= (
i
4
)
5
*
i
= (1)
5
*
i
=
i
2. Solve for
x
:
(a)
3 + 2

x
+ 4

=
0
2

x
+ 4

=

3

x
+ 4

=

3
2
Since

x
 ≥
0 for all
x
∈
R
then

x
+ 4

=

3
2
has no solution. So
the solution set is
∅
.
(b)
3 +
√
3
x
+ 1
=
x
(1)
√
3
x
+ 1
=
x

3
(2)
(
√
3
x
+ 1)
2
=
(
x

3)
2
(3)
3
x
+ 1
=
x
2

6
x
+ 9
(4)
0
=
x
2

9
x
+ 8
(5)
0
=
(
x

8)(
x

1)
(6)
So
x
= 8 and
x
= 1. However in step (3) we could have created
extranenous solutions. Thus, we must check our solutions.
1
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Check
x
= 8:
3 +
√
3
*
8 + 1 = 3 +
√
25 = 8. So 8 works.
Check
x
= 1:
3 +
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 Fall '08
 WILLIAMSON
 Math, Calculus, Algebra, Equations, Quadratic equation, complex conjugates, Quiz 6A

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