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Math 20-1

Math 20-1 · Worksheets

Mixed review · Radicals and rational expressions

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

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Mixed review · Radicals and rational expressions

Math 20-1 · maddyhelps.com

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

  1. The solution to √(x + 7) = 4 is

    1. a) x = 16
    2. b) x = 23
    3. c) x = −3
    4. d) x = 9

  2. The solution to 3/(x − 2) = x is

    1. a) no solution
    2. b) x = 3 only
    3. c) x = 3 or x = −1
    4. d) x = 2 or x = −1

  3. Rationalizing the denominator of 6/√3 gives

    1. a) √2
    2. b) 6√3
    3. c) 2√3
    4. d) 2√3/3

  4. The non-permissible values of x/(x² − 4) are

    1. a) x = 2 and x = −2
    2. b) x = 2 only
    3. c) x = 4
    4. d) x = 0

  5. (x² − 1)/(x + 2) ÷ (x − 1)/(x + 2) equals

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

  6. √18 + √8 simplifies to

    1. a) 5√2
    2. b) 26√2
    3. c) √26
    4. d) 2√13

  7. (x² − 9)/(x + 3) simplifies to

    1. a) x − 9, x ≠ −3
    2. b) x + 3, x ≠ 3
    3. c) x − 3, x ≠ −3
    4. d) x − 3

  8. The solution to √(2x + 3) = x is

    1. a) x = 3 only
    2. b) no solution
    3. c) x = 3 or x = −1
    4. d) x = −1 only

  9. The non-permissible value of (x + 2)/(x − 5) is

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

  10. (x² + 5x + 6)/(x² − 4) simplifies to

    1. a) (5x + 6)/(−4)
    2. b) (x + 3)/(x − 2), x ≠ ±2
    3. c) (x + 3)/(x − 2)
    4. d) (x + 3)/(x + 2), x ≠ ±2

Answer key · Mixed review · Radicals and rational expressions

Math 20-1 · maddyhelps.com

  1. d) x = 9 — Square both sides: x + 7 = 16, so x = 9. Check it in the original: √16 = 4 ✓. Squaring is safe here, but always check, because squaring can create roots the original equation never had.
  2. c) x = 3 or x = −1 — Multiply both sides by (x − 2): 3 = x² − 2x, so x² − 2x − 3 = 0 and (x − 3)(x + 1) = 0. Both x = 3 and x = −1 are allowed, since neither is the non-permissible value x = 2.
  3. c) 2√3 — Multiply top and bottom by √3: (6√3)/(√3 · √3) = 6√3/3 = 2√3. Multiplying by √3/√3 is multiplying by 1, so the value does not change — only its appearance does.
  4. a) x = 2 and x = −2 — Factor the denominator: x² − 4 = (x − 2)(x + 2), which is zero at x = 2 and at x = −2. A quadratic denominator usually gives two restrictions, so factor before you look.
  5. b) x + 1 — Dividing is multiplying by the reciprocal: [(x − 1)(x + 1)/(x + 2)] · [(x + 2)/(x − 1)]. The (x + 2) and the (x − 1) both cancel, leaving x + 1.
  6. a) 5√2 — Radicals only add when they are like radicals, so simplify first: √18 = 3√2 and √8 = 2√2. Then 3√2 + 2√2 = 5√2. You cannot add the numbers under the roots — √18 + √8 is not √26.
  7. c) x − 3, x ≠ −3 — Factor first: (x − 3)(x + 3)/(x + 3), and the (x + 3) cancels to leave x − 3. The restriction stays: the original was undefined at x = −3, and cancelling does not change that, so it must be written down.
  8. a) x = 3 only — Squaring gives 2x + 3 = x², so x² − 2x − 3 = 0 and (x − 3)(x + 1) = 0, giving x = 3 or x = −1. Check both: √9 = 3 ✓, but √1 = 1, not −1, so x = −1 is extraneous. It appeared because squaring erased the sign.
  9. d) x = 5 — A rational expression is undefined when its denominator is zero. Set x − 5 = 0 to get x = 5. The numerator being zero is fine — it just makes the whole expression zero.
  10. b) (x + 3)/(x − 2), x ≠ ±2 — Factor both: (x + 3)(x + 2)/[(x − 2)(x + 2)]. The (x + 2) cancels, leaving (x + 3)/(x − 2). Both original restrictions survive — x ≠ 2 and x ≠ −2 — even though only one of them is still visible.