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Math 20-1

Math 20-1 · Worksheets

Final exam challenge

Ten questions of mixed difficulty, covering Inequalities and graphs, Quadratic equations, Radicals, Rational expressions, Trigonometry. Print it, or work through it on screen — the answer key starts on its own page.

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Final exam challenge

Math 20-1 · maddyhelps.com

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Circle the best answer for each question. Show your work in the space provided.

  1. For √(x − 5) to be a real number, x must satisfy

    1. a) x ≥ 0
    2. b) x > 5
    3. c) x ≥ 5
    4. d) x ≤ 5

  2. Factored completely, (x + 1)² − 9 is

    1. a) (x + 4)(x − 2)
    2. b) (x + 1)(x − 9)
    3. c) (x + 1 − 3)²
    4. d) (x + 10)(x − 8)

  3. Compared with y = 2x − 6, the graph of y = |2x − 6|

    1. a) is identical everywhere
    2. b) is identical for x ≥ 3 and reflected in the x-axis for x < 3
    3. c) is the whole line reflected in the x-axis
    4. d) is the whole line shifted up

  4. The solution to x² − 5x + 6 = 0 is

    1. a) x = 1 or x = 6
    2. b) x = −2 or x = −3
    3. c) x = 5 or x = 6
    4. d) x = 2 or x = 3

  5. Rationalizing the denominator of 4/(3 − √5) gives

    1. a) 3 + √5
    2. b) 4(3 − √5)/4
    3. c) (3 + √5)/4
    4. d) 3 − √5

  6. (x/3) · (6/x²) equals

    1. a) 2/x, x ≠ 0
    2. b) 2/x²
    3. c) 6x/(3x²)
    4. d) 2x

  7. In triangle ABC, a = 8, ∠A = 40° and ∠B = 65°. The length of b is about

    1. a) 9.4
    2. b) 5.7
    3. c) 11.3
    4. d) 12.7

  8. The inequality x² + 1 < 0 has

    1. a) no solution
    2. b) x = 0 as its only solution
    3. c) all real numbers as solutions
    4. d) x < −1 as its solution

  9. Factored, x² − 16 is

    1. a) (x − 4)(x + 4)
    2. b) (x − 8)(x + 2)
    3. c) (x² − 4)(x + 4)
    4. d) (x − 4)²

  10. Factored, 2x² + 7x + 3 is

    1. a) 2(x + 1)(x + 3)
    2. b) (2x + 7)(x + 3)
    3. c) (2x + 3)(x + 1)
    4. d) (2x + 1)(x + 3)

Answer key · Final exam challenge

Math 20-1 · maddyhelps.com

  1. c) x ≥ 5 — A square root of a negative number is not real, so the radicand must be zero or more: x − 5 ≥ 0, giving x ≥ 5. Five itself is allowed, because √0 = 0.
  2. a) (x + 4)(x − 2) — Treat it as a difference of squares with a = x + 1 and b = 3: (x + 1 − 3)(x + 1 + 3) = (x − 2)(x + 4). Expanding first also works and gives x² + 2x − 8, which factors the same way.
  3. b) is identical for x ≥ 3 and reflected in the x-axis for x < 3 — Absolute value leaves non-negative outputs alone and flips negative ones. The line is negative to the left of its x-intercept x = 3, so only that part is reflected, producing a corner at (3, 0).
  4. d) x = 2 or x = 3 — Two numbers multiply to 6 and add to −5: −2 and −3. So (x − 2)(x − 3) = 0 and the roots are 2 and 3. The roots come out with the opposite signs to the numbers in the brackets.
  5. a) 3 + √5 — Multiply top and bottom by the conjugate 3 + √5. The denominator becomes 3² − (√5)² = 9 − 5 = 4, so you get 4(3 + √5)/4 = 3 + √5. The conjugate works because the middle terms cancel, leaving no radical.
  6. a) 2/x, x ≠ 0 — Multiply straight across: 6x/(3x²), then reduce to 2/x. Cancelling before multiplying is faster and gives the same thing: the 3 into the 6 and one x from x².
  7. c) 11.3 — Two angles and the side opposite one of them means the sine law: b/sin B = a/sin A, so b = 8 sin 65°/sin 40° ≈ 8(0.906)/0.643 ≈ 11.3. B is the larger angle, so b should be the longer side, which it is.
  8. a) no solution — A square is never negative, so x² + 1 is at least 1 for every real x and can never be below zero. The discriminant agrees: 0 − 4 = −4, so the parabola never touches the axis and sits entirely above it.
  9. a) (x − 4)(x + 4) — This is a difference of squares, a² − b² = (a − b)(a + b), with a = x and b = 4. A difference of squares always splits; a sum of squares like x² + 16 does not factor over the real numbers.
  10. d) (2x + 1)(x + 3) — With a leading coefficient, check by expanding. (2x + 1)(x + 3) gives 2x² + 6x + x + 3 = 2x² + 7x + 3 ✓. The near-miss (2x + 3)(x + 1) gives a middle term of 5x, so the order of the 1 and the 3 matters.