Mathematics 133 – Intensive Calculus for Science 1 “Quiz 0” – Algebra Review September 2, 2005

Your Name: \underline{Solutions − −AnswerKey}

Do all work on this sheet (front and back). You have 20 minutes. No calculators on this quiz.

A) Write as a single quotient and simplify: \frac{\frac{1}{x}}{3−x^{2}} + \frac{2−x}{3+x^{2}}

To add fractions, we find a common denominator, then add the numerators:

\frac{a}{b} + \frac{c}{d} = \frac{ad}{bd} + \frac{bc}{bd} = \frac{ad + bc}{bd}

Here the result is

\frac{\frac{1}{x}}{3 − x^{2}} + \frac{2 − x}{3 + x^{2}} = \frac{\frac{1}{x}(3 + x^{2})}{(3 − x^{2})(3 + x^{2})} + \frac{(2 − x)(3 − x^{2})}{(3 + x^{2})(3 − x^{2})}

= \frac{\frac{3}{x} + x + 6 − 3x − 2x^{2} + x^{3}}{9 − x^{4}}

= \frac{3 + 6x − 2x^{2} − 2x^{3} + x^{4}}{9x − x^{5}}

It’s also OK to leave the denominator in factored form: x(3 − x^{2})(3 + x^{2}).

B) Solve for t: 5(t − 2) + 1 = 12 − t.

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.

C) Solve for x: 2x^{2} − 3x = 3 (find all solutions)

The left side of the equation 2x^{2} − 3x − 3 = 0 does not factor easily, so we use the Quadratic Formula: The roots of the quadratic equation ax^{2} + bx + c = 0 are

x = \frac{−b ± \sqrt{b^{2} − 4ac}}{2a}.

Here a = 2, b = −3, c = −3, so

x = \frac{3 ± \sqrt{9 + 24}}{4} = \frac{3 ± \sqrt{33}}{4}

D) Solve for a: 27a^{1/3} = 3a First method: Cube both sides to get rid of the cube root a^{1/3} = \sqrt[3]{a}: 27^{3}a = 27a^{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.

Second method: Notice that a = 0 is one solution. If a ≠ 0, then we can divide both sides by 3a^{1/3} to get 9 = a^{2/3}. Hence a = 9^{3/2} = (±\sqrt{9})^{3} = (±3)^{3} = ±27.

E) Simplify: ((1 + c)^{3/2})^{4} · (1 + c)^{−2}

By the rules for exponents:

((1 + c)^{3/2})^{4} · (1 + c)^{−2} = (1 + c)^{4·3/2}(1 + c)^{−2} = (1 + c)^{6−2} = (1 + c)^{4}.

F) Expand and simplify: (a^{2} − 2b + 2)(2b − a + 1)

Answer: 2a^{2}b−4b^{2}+4b−a^{3}+2ab−2a+a^{2}−2b+2 = 2a^{2}b−4b^{2}+2b−a^{3}+2ab−2a+a^{2}+2.

G) Find the coordinates of the center and the radius of the circle with equation x^{2} − 6x + y^{2} + 4y = 0.

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}.