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\leftline{Mathematics 133 -- Intensive Calculus for Science 1}
\leftline{``Quiz 0'' -- Algebra Review September 2, 2005}
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\leftline{Your Name: $\underline{Solutions -- Answer Key}$}
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Do all work on this sheet (front and back).  You have 20 minutes. {\bf No calculators on this quiz.}
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A) Write as a single quotient and simplify:  ${{1\over x}\over 3 - x^2} + {2 - x\over 3+x^2}$

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To add fractions, we find a common denominator, then add the numerators:
$${a\over b} + {c\over d} = {ad\over bd} + {bc\over bd} = {ad + bc\over bd}$$
Here the result is
$$\eqalign{ {{1\over x}\over 3-x^2} + {2 - x\over 3+x^2}
&={{1\over x}(3+x^2) \over (3-x^2)(3+x^2)} + {(2 - x)(3-x^2)\over (3+x^2)(3-x^2)}\cr
&= {{3\over x} + x + 6 - 3x - 2x^2 +x^3\over 9 - x^4}\cr
&= {3 + 6x - 2x^2 - 2 x^3 + x^4\over 9x - x^5}\cr}$$
It's also OK to leave the denominator in factored form:  $x(3-x^2)(3+x^2)$.
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B) Solve for $t$:  $5(t - 2) + 1 = 12 - t$.
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\item{} Multiply out on the left:  $5t - 10 + 1 = 12 - t$ then add $t + 9$
to both sides to put all the terms with $t$ on the left:  $6t = 21$.  
Hence $t = 21/6 = 7/2$.
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C) Solve for $x$: $2x^2 - 3x = 3$ (find all solutions)
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\item{}  The left side of the equation $2x^2 - 3x - 3 = 0$ does not factor
easily, so we use the {\it Quadratic Formula}: The roots of the quadratic
equation $a x^2 + bx + c = 0$ are 
$$x = {-b \pm \sqrt{b^2 - 4ac}\over 2a}.$$
Here $a = 2, b = -3, c = -3$, so 
$$x = {3 \pm \sqrt{9 + 24}\over 4} = {3 \pm \sqrt{33}\over 4}$$
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D) Solve for $a$:  $27 a^{1/3} = 3 a$
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\item{} {\it First method}:  Cube both sides to get rid of the cube root
$a^{1/3} = {\root 3\of a}$:  $27^3 a = 27 a^3$, or $a^3 - 27^2 a = 0$.
This factors as $a(a - 27)(a + 27) = 0$, so $a = 0$, $a = 27$, or $a = -27$.
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\item{} {\it Second method}: Notice that $a = 0$ is one solution.
If $a \ne 0$, then we can divide both sides by $3a^{1/3}$ to get
$9 = a^{2/3}$.  Hence $a = 9^{3/2} = (\pm \sqrt{9})^3 = (\pm 3)^3 = \pm 27$.
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E)  Simplify:  $\left((1 + c)^{3/2}\right)^4 \cdot (1 + c)^{-2}$
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\item{}  By the rules for exponents:
$$\left((1 + c)^{3/2}\right)^4 \cdot (1 + c)^{-2} = (1+c)^{4\cdot 3/2}(1+c)^{-2} 
= (1+c)^{6-2} = (1+c)^4.$$
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F) Expand and simplify:  $(a^2 - 2b + 2)(2b - a + 1)$
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\item{} Answer:  $2a^2b - 4b^2 + 4b - a^3 + 2ab - 2a + a^2 - 2b + 2 = 
2a^2b - 4b^2 + 2b - a^3 + 2ab - 2a + a^2 + 2$.
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G)  Find the coordinates of the center and the radius
of the circle with equation $x^2 - 6x + y^2 + 4y = 0$.
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\item{} Answer: Completing the square in $x$ and $y$, we find:
$(x-3)^2 - 9 + (y+2)^2 - 4 = 0$, or $(x-3)^2 + (y+2)^2 = 13$.
The center is at $(3,-2)$ and the radius is $\sqrt{13}$.
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