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\leftline{\bf  Math 131 - section 01 -  Precalculus Diagnostic Quiz Answers}
\leftline{\bf  August 31,  2007}

\vspace{.2in}
\noindent
Circle the correct answer. {\bf Show your work.}  Please turn over for problems 4,5,6.
\begin{enumerate}
\item Simplify: $\displaystyle \sqrt{15}\left(\frac{1}{\sqrt{3}}+\frac{1}{\sqrt{5}}\right)$\vspace{.1in}

A. \ $\displaystyle 2 \frac{\sqrt{15}}{\sqrt{8}}$\ \ \ \ \ B. \ $\displaystyle \frac{\sqrt{15}}{\sqrt{3}+\sqrt{5}}$ \ \ \ \ \ C. \ $1$ \ \ \ \ \ D. \ $\displaystyle \sqrt{3}+\sqrt{5}$\ \ \ \ \ E. \ $\displaystyle 2 \sqrt{3}$.

\begin{itemize}
\item {\it Solutions:}  The answer is D.  There are several ways to do this, but all rely
on the fact that $\sqrt{ab} = \sqrt{a}\cdot\sqrt{b}$ for all positive real $a,b$.

\item Method 1:  Multiply through:
\begin{eqnarray*}
\sqrt{15}\left(\frac{1}{\sqrt{3}} + \frac{1}{\sqrt{5}}\right) &=& 
\frac{\sqrt{15}}{\sqrt{3}} + \frac{\sqrt{15}}{\sqrt{5}} \\
&=& \frac{\sqrt{5}\sqrt{3}}{\sqrt{3}} + \frac{\sqrt{5}\sqrt{3}}{\sqrt{5}}\\
&=& \sqrt{5} + \sqrt{3}
\end{eqnarray*}

\item Method 2: Put the terms on the inside the parentheses over a common denominator:
$$\frac{1}{\sqrt{3}} + \frac{1}{\sqrt{5}} = \frac{\sqrt{5} + \sqrt{3}}{\sqrt{15}}$$
Then multiplying by $\sqrt{15}$ gives $\sqrt{5} + \sqrt{3}$ again.
\end{itemize}

\item Simplify: $\displaystyle \frac{12x}{3x-6} \cdot \frac{x^2-4}{2x+4}$\vspace{.1in}

A. \ $\displaystyle \frac{x^2+12x-4}{5x-2}$ \ \ \ \ \ B. \ $\displaystyle 2x$ \ \ \ \ \ C. \ $\displaystyle -\frac{4}{3}(x-1)$ \ \ \ \ \ D. \ $\displaystyle \frac{2x(x-2)}{x+2}$ \ \ \ \ \ E. \ $\displaystyle \frac{12x^3-48x}{6x^2-12}$

\begin{itemize}
\item {\it Solution:}
The answer is B. Recall the difference of squares factorization:  $x^2 - a^2 = (x + a)(x - a)$.
Here, $x^2 - 4 = (x + 2)(x - 2)$, so 
$$\frac{12x}{3x-6} \cdot \frac{x^2-4}{2x+4} = \frac{12 x (x + 2)(x - 2)}{3(x-2)\cdot 2(x + 2)}$$
Cancelling common factors top and bottom yields $2x$.
\end{itemize}

\vspace{.1in}

\item Simplify: $\displaystyle \left(u^{-5}v^2 \right)^3 \left(\frac{v^2}{u}\right)^{-1}$\vspace{.1in}

A. \ $u^{-16}v^8$ \ \ \ \ \ B. \ $u^{-14}v^{4}$ \ \ \ \ \ C. \ $u^{-9}v^{3}$ \ \ \ \ \ D. \ $(uv)^{-7}$ \ \ \ \ \ E. \  $u^{-4}v^{2}$

\begin{itemize}
\item {\it Solution:} The answer is B.  Recall the rules for exponents:
$$a^b\cdot a^c = a^{b + c}, \qquad (a^b)^c = a^{b \cdot c}, \qquad a^{-1} = \frac{1}{a}.$$
Then 
$$\left(u^{-5}v^2 \right)^3 \left(\frac{v^2}{u}\right)^{-1}
= u^{-15} v^6 \cdot u v^{-2} = u^{-14} v^4.$$
\end{itemize}

\item If $f(x) = 5x^2-11$, what is $f(a-2)$? 

\begin{itemize}
\item {\it Solution:} Substitute $a - 2$ for the $x$ in the formula to yield
$$f(a-2) = 5(a-2)^2 - 11 = 5(a^2 - 4a + 4) - 11 = 5a^2 - 20a + 9.$$
\end{itemize}

\item Find all values of $x$ satisfying $2(x-2)>5$. 

\begin{itemize}
\item {\it Solution:} All real $x > 9/2$ (since $x - 2 > 5/2$, so $x > 5/2 + 2 = 9/2$).  
\end{itemize}

\item Solve for $x$:  \ \ $2x^2 - x - 6=0$ (find all solutions). 

\begin{itemize}
\item {\it Solution:}  This quadratic factors as $(2x + 3)(x - 2) = 0$, so
$2x + 3 = 0$ or $x - 2 = 0$.  Hence $x = -3/2$ or $x = 2$.  The equation 
can also be solved by the quadratic formula:
$$x = \frac{1 \pm \sqrt{1 + 49}}{4} = 2, -3/2.$$ 
\end{itemize}


\end{enumerate}

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