Math 131 - section 01 - Precalculus Diagnostic Quiz Answers August 31, 2007

Circle the correct answer. Show your work. Please turn over for problems 4,5,6.

1. Simplify: \sqrt{15} (\frac{1}{\sqrt{3}} + \frac{1}{\sqrt{5}})

A. 2\begin{matrix}√ & \overline{15} \\ √ & \overline{8} \end{matrix} B. \frac{\sqrt{15}}{\sqrt{3} + \sqrt{5}} C. 1 D. \sqrt{3} + \sqrt{5} E. 2\sqrt{3}.

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

• Method 1: Multiply through:

\sqrt{15} (\frac{1}{\sqrt{3}} + \frac{1}{\sqrt{5}}) = \begin{matrix}√ & \overline{15} \\ √ & \overline{3} \end{matrix} + \begin{matrix}√ & \overline{15} \\ √ & \overline{5} \end{matrix}
= \begin{matrix}\sqrt{5} & \sqrt{3} \\ √ & \overline{3} \end{matrix} + \begin{matrix}\sqrt{5} & \sqrt{3} \\ √ & \overline{5} \end{matrix}
= \sqrt{5} + \sqrt{3}

• 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.

2. Simplify: \frac{12x}{3x − 6} · \frac{x^{2} − 4}{2x + 4}

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

• 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} · \frac{x^{2} − 4}{2x + 4} = \frac{12x(x + 2)(x − 2)}{3(x − 2) · 2(x + 2)}

Cancelling common factors top and bottom yields 2x.

3. Simplify: (u^{−5}v^{2})^{3} (\frac{v^{2}}{u} )^{−1}

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

• Solution: The answer is B. Recall the rules for exponents:

a^{b} · a^{c} = a^{b+c},  (a^{b})^{c} = a^{b·c},  a^{−1} = \frac{1}{a}.

Then

(u^{−5}v^{2})^{3} (\frac{v^{2}}{u} )^{−1} = u^{−15}v^{6} · uv^{−2} = u^{−14}v^{4}.

4. If f(x) = 5x^{2} − 11, what is f(a − 2)?

• 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.

5. Find all values of x satisfying 2(x − 2) > 5.

• Solution: All real x > 9/2 (since x − 2 > 5/2, so x > 5/2 + 2 = 9/2).

6. Solve for x: 2x^{2} − x − 6 = 0 (find all solutions).

• 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 ± \sqrt{1 + 49}}{4} = 2, −3/2.