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Unformatted text preview: Solutions to Quiz 6A
www.math.uﬂ.edu/˜harringt
October 5, 2007
1. Reduce and simplify
(a) (3+2i)5
(1
7i + i2 ) = (3+2i)5
(1
7i − 1) = (3+2i)5
7i ∗0=0 (b) i21 = (i4 )5 ∗ i = (1)5 ∗ i = i 2. Solve for x:
(a)
3 + 2x + 4 = 0
2x + 4 = −3
−3
x + 4 =
2
Since x ≥ 0 for all x ∈ R then x + 4 =
the solution set is ∅. −3
2 has no solution. So (b)
√
3x + 1
√
3x + 1
√
( 3x + 1)2
3x + 1
0
0 3+ =
=
=
=
=
= x
x−3
(x − 3)2
x2 − 6x + 9
x2 − 9x + 8
(x − 8)(x − 1) (1)
(2)
(3)
(4)
(5)
(6) So x = 8 and x = 1. However in step (3) we could have created
extranenous solutions. Thus, we must check our solutions. 1 Check x = 8:
3+ √ 3∗8+1=3+ √ 25 = 8. So 8 works. Check x = 1:
3+ √ 3∗1+1=3+ √ 4 = 5 = 1. So 1 does NOT work. So the solution set is {8}.
3. Determine whether the statement is true or false. (a) If 3 + 2i is a root to a quadratic equation with real coeﬃcients
then the other root is 3 − 2i.
This statement is TRUE. The roots are complex conjugates of
each other.
√
√√
(b) −3 −2 = 6
√√
√
√√
This statement is FALSE. −3 −2 = i2 3 2 = − 6. 2 ...
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This note was uploaded on 07/12/2011 for the course MAC 1140 taught by Professor Williamson during the Fall '08 term at University of Florida.
 Fall '08
 WILLIAMSON
 Math, Calculus, Algebra

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