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Diagnostic quiz

30 questions · about 40-50 minutes · no time limit

This quiz finds out what you already know, so you can skip it and spend your time where it counts. The questions go in course order: Unit 1 first (easiest), Unit 11 last (hardest).

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Compute \(\dfrac{2}{3} - \dfrac{5}{6}\).

The LCD of 3 and 6 is 6. Rewrite: \(\dfrac{2}{3} = \dfrac{4}{6}\).

\(\dfrac{4}{6} - \dfrac{5}{6} = \dfrac{4 - 5}{6} = \dfrac{-1}{6}\).

Answer: \(-\dfrac{1}{6}\).

Simplify \(5(x - 2) - 3(2x - 1)\).

Distribute the 5 and the \(-3\): \(5x - 10 - 6x + 3\). (The \(-3\) times \(-1\) gives \(+3\).)

Combine like terms: \(5x - 6x = -x\) and \(-10 + 3 = -7\).

Answer: \(-x - 7\).

Solve \(4(x - 3) = 2x + 6\).

Distribute: \(4x - 12 = 2x + 6\).

Subtract \(2x\) from both sides: \(2x - 12 = 6\). Add 12: \(2x = 18\). Divide by 2: \(x = 9\).

Check: \(4(9 - 3) = 24\) and \(2(9) + 6 = 24\). Answer: \(x = 9\).

Solve \(-5 \le 3x - 2 \lt 7\). Give the answer in interval notation.

Add 2 to all three parts: \(-3 \le 3x \lt 9\).

Divide all three parts by 3 (positive, so no flip): \(-1 \le x \lt 3\).

\(-1\) is included (bracket), 3 is not (parenthesis). Answer: \([-1, 3)\).

Solve \(|2x - 3| = 7\). List every solution.

The inside is either 7 or \(-7\).

Case 1: \(2x - 3 = 7\), so \(2x = 10\) and \(x = 5\).

Case 2: \(2x - 3 = -7\), so \(2x = -4\) and \(x = -2\).

Answer: \(x = 5\) or \(x = -2\).

Find the slope of the line through \((-2, 5)\) and \((4, -1)\).

\(m = \dfrac{y_2 - y_1}{x_2 - x_1} = \dfrac{-1 - 5}{4 - (-2)} = \dfrac{-6}{6}\).

Answer: \(m = -1\).

Write the equation of the line through \((2, 1)\) that is perpendicular to \(y = 2x - 3\). Give it in the form \(y = mx + b\).

The given line has slope 2. A perpendicular slope is the opposite reciprocal: \(-\tfrac{1}{2}\).

Point-slope form: \(y - 1 = -\tfrac{1}{2}(x - 2)\), so \(y - 1 = -\tfrac{1}{2}x + 1\).

Add 1: \(y = -\tfrac{1}{2}x + 2\).

Solve the system: \(2x + 3y = 7\) and \(x - 3y = 8\). Give the answer as a point \((x, y)\).

Add the equations (the \(3y\) and \(-3y\) cancel): \(3x = 15\), so \(x = 5\).

Put \(x = 5\) into \(x - 3y = 8\): \(5 - 3y = 8\), so \(-3y = 3\) and \(y = -1\).

Check in the first equation: \(2(5) + 3(-1) = 7\). Answer: \((5, -1)\).

Adult tickets cost $9 and child tickets cost $5. A group bought 30 tickets for $206 in total. How many adult tickets did they buy?

Let \(a\) = adult tickets and \(c\) = child tickets. Then \(a + c = 30\) and \(9a + 5c = 206\).

Substitute \(c = 30 - a\): \(9a + 5(30 - a) = 206\), so \(4a + 150 = 206\).

\(4a = 56\), so \(a = 14\) (and \(c = 16\)). Check: \(9 \cdot 14 + 5 \cdot 16 = 126 + 80 = 206\).

Answer: 14 adult tickets.

Simplify \(\dfrac{(2x^3y^{-2})^2}{4x^{-1}y}\). Write the answer with positive exponents only.

Power rule on top: \((2x^3y^{-2})^2 = 4x^6y^{-4}\).

Divide: \(\dfrac{4}{4} = 1\), \(x^{6 - (-1)} = x^7\), \(y^{-4 - 1} = y^{-5}\).

Move \(y^{-5}\) to the bottom: answer \(\dfrac{x^7}{y^5}\).

Simplify \((x + 3)(x - 3) - (x - 2)^2\).

\((x + 3)(x - 3) = x^2 - 9\) and \((x - 2)^2 = x^2 - 4x + 4\).

Subtract the whole second product: \(x^2 - 9 - (x^2 - 4x + 4) = x^2 - 9 - x^2 + 4x - 4\).

Combine like terms. Answer: \(4x - 13\).

Divide: \((6x^2 + x - 12) \div (2x + 3)\).

\(6x^2 \div 2x = 3x\). Multiply: \(3x(2x + 3) = 6x^2 + 9x\). Subtract: \(x - 9x = -8x\); bring down \(-12\).

\(-8x \div 2x = -4\). Multiply: \(-4(2x + 3) = -8x - 12\). Subtract: remainder 0.

Answer: \(3x - 4\). Check: \((3x - 4)(2x + 3) = 6x^2 + 9x - 8x - 12 = 6x^2 + x - 12\).

Factor completely: \(2x^3 - 8x^2 - 24x\).

GCF first: \(2x^3 - 8x^2 - 24x = 2x(x^2 - 4x - 12)\).

Two numbers that multiply to \(-12\) and add to \(-4\): \(-6\) and \(2\). So \(x^2 - 4x - 12 = (x - 6)(x + 2)\).

Answer: \(2x(x - 6)(x + 2)\).

Factor: \(6x^2 + 7x - 3\).

\(a \cdot c = 6 \cdot (-3) = -18\). Two numbers that multiply to \(-18\) and add to 7: \(9\) and \(-2\).

Split the middle: \(6x^2 + 9x - 2x - 3 = 3x(2x + 3) - 1(2x + 3)\).

Answer: \((3x - 1)(2x + 3)\).

Solve \(x^2 - 5x = 14\). List every solution.

Standard form first: \(x^2 - 5x - 14 = 0\).

Factor: \((x - 7)(x + 2) = 0\).

Zero product property: \(x - 7 = 0\) or \(x + 2 = 0\). Answer: \(x = 7\) or \(x = -2\).

Simplify \(\dfrac{x^2 - 9}{x^2 + x - 6}\).

Factor top and bottom: \(\dfrac{(x - 3)(x + 3)}{(x + 3)(x - 2)}\).

Cancel the common factor \(x + 3\). Answer: \(\dfrac{x - 3}{x - 2}\).

Subtract and simplify: \(\dfrac{x}{x - 2} - \dfrac{8}{x^2 - 4}\).

\(x^2 - 4 = (x - 2)(x + 2)\), so the LCD is \((x - 2)(x + 2)\).

Rewrite the first fraction: \(\dfrac{x(x + 2)}{(x - 2)(x + 2)} - \dfrac{8}{(x - 2)(x + 2)} = \dfrac{x^2 + 2x - 8}{(x - 2)(x + 2)}\).

Factor the top: \(x^2 + 2x - 8 = (x + 4)(x - 2)\). Cancel \(x - 2\).

Answer: \(\dfrac{x + 4}{x + 2}\).

Solve \(\dfrac{3}{x} + \dfrac{x}{x + 2} = \dfrac{-2}{x + 2}\). List every solution (or type none).

Multiply every term by the LCD \(x(x + 2)\): \(3(x + 2) + x \cdot x = -2x\).

So \(x^2 + 3x + 6 = -2x\), which gives \(x^2 + 5x + 6 = 0\), or \((x + 2)(x + 3) = 0\).

\(x = -2\) makes the denominator \(x + 2\) equal 0, so it is extraneous. Reject it.

Answer: \(x = -3\).

Simplify \((8x^6)^{2/3}\).

Apply the exponent to each factor: \(8^{2/3} \cdot x^{6 \cdot \frac{2}{3}}\).

\(8^{2/3} = (\sqrt[3]{8})^2 = 2^2 = 4\), and \(6 \cdot \tfrac{2}{3} = 4\).

Answer: \(4x^4\).

Rationalize the denominator and simplify: \(\dfrac{4}{3 - \sqrt{5}}\).

  1. \(3 - \sqrt{5}\)
  2. \(3 + \sqrt{5}\)
  3. \(-3 - \sqrt{5}\)
  4. \(12 + 4\sqrt{5}\)

Multiply top and bottom by the conjugate \(3 + \sqrt{5}\).

Bottom: \((3 - \sqrt{5})(3 + \sqrt{5}) = 9 - 5 = 4\). Top: \(4(3 + \sqrt{5})\).

\(\dfrac{4(3 + \sqrt{5})}{4} = 3 + \sqrt{5}\). Answer: b.

Solve \(\sqrt{2x + 3} = x\). List every solution (or type none).

Square both sides: \(2x + 3 = x^2\), so \(x^2 - 2x - 3 = 0\), or \((x - 3)(x + 1) = 0\).

Check \(x = 3\): \(\sqrt{9} = 3\). Works.

Check \(x = -1\): \(\sqrt{1} = 1\), not \(-1\). Extraneous, reject it.

Answer: \(x = 3\).

Multiply and write in the form \(a + bi\): \((3 + 2i)(1 - 4i)\).

FOIL: \(3 - 12i + 2i - 8i^2\).

Use \(i^2 = -1\): \(-8i^2 = +8\). So \(3 + 8 - 10i\).

Answer: \(11 - 10i\).

Solve \(x^2 - 4x - 1 = 0\). Give exact answers.

It does not factor, so use the quadratic formula with \(a = 1\), \(b = -4\), \(c = -1\).

\(x = \dfrac{4 \pm \sqrt{(-4)^2 - 4(1)(-1)}}{2} = \dfrac{4 \pm \sqrt{20}}{2}\).

\(\sqrt{20} = 2\sqrt{5}\), so \(x = \dfrac{4 \pm 2\sqrt{5}}{2} = 2 \pm \sqrt{5}\).

Find the vertex of the parabola \(y = 2x^2 - 8x + 3\). Give it as a point.

\(x = -\dfrac{b}{2a} = -\dfrac{-8}{2 \cdot 2} = 2\).

\(y = 2(2)^2 - 8(2) + 3 = 8 - 16 + 3 = -5\).

Answer: \((2, -5)\).

Solve \(x^2 - x - 6 \le 0\). Give the answer in interval notation.

Find the zeros: \(x^2 - x - 6 = (x - 3)(x + 2) = 0\), so \(x = 3\) or \(x = -2\).

The parabola opens up, so it is at or below 0 between the zeros. (Test \(x = 0\): \(-6 \le 0\), true.)

The zeros count because of \(\le\). Answer: \([-2, 3]\).

Find the domain of \(f(x) = \dfrac{\sqrt{x - 4}}{x - 6}\). Give it in interval notation.

The square root needs \(x - 4 \ge 0\), so \(x \ge 4\).

The denominator cannot be 0, so \(x \ne 6\).

Answer: \([4, 6) \cup (6, \infty)\).

Find the inverse of \(f(x) = \dfrac{2x - 1}{3}\). What is \(f^{-1}(x)\)?

Write \(y = \dfrac{2x - 1}{3}\), then swap \(x\) and \(y\): \(x = \dfrac{2y - 1}{3}\).

Solve for \(y\): \(3x = 2y - 1\), so \(2y = 3x + 1\) and \(y = \dfrac{3x + 1}{2}\).

Answer: \(f^{-1}(x) = \dfrac{3x + 1}{2}\).

Evaluate \(\log_4 8\).

\(\log_4 8 = y\) means \(4^y = 8\).

Write both with base 2: \(2^{2y} = 2^3\), so \(2y = 3\).

Answer: \(y = \dfrac{3}{2}\).

Solve \(5^x = 12\). Give the exact answer.

Take the natural log of both sides: \(\ln(5^x) = \ln 12\).

Power property: \(x \ln 5 = \ln 12\).

Answer: \(x = \dfrac{\ln 12}{\ln 5}\) (about 1.544).

Solve \(\log_2 x + \log_2 (x - 2) = 3\). List every solution (or type none).

Combine with the product property: \(\log_2 \big(x(x - 2)\big) = 3\).

Exponential form: \(x(x - 2) = 2^3 = 8\), so \(x^2 - 2x - 8 = 0\), or \((x - 4)(x + 2) = 0\).

\(x = -2\) would need \(\log_2(-2)\), which is undefined. Reject it.

Answer: \(x = 4\).

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