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Math 20-1

Math 20-1 · Worksheets

Absolute value and radicals

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

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Absolute value and radicals

Math 20-1 · maddyhelps.com

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

  1. Rationalizing the denominator of 6/√3 gives

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

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

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

  3. The value of |−7 + 3| is

    1. a) −4
    2. b) 4
    3. c) −10
    4. d) 10

  4. The solution to √(x + 7) = 4 is

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

  5. The solution to |2x − 5| = 9 is

    1. a) x = 7 or x = −2
    2. b) x = 2 or x = −7
    3. c) x = 7 only
    4. d) x = 2 only

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

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

  7. If x < 0, then |x| is equal to

    1. a) 0
    2. b) x
    3. c) −x
    4. d) x²

  8. The product (2√3)(5√6) equals

    1. a) 10√18
    2. b) 10√2
    3. c) 30√2
    4. d) 30√18

  9. √18 + √8 simplifies to

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

  10. The solution to |x − 3| = 2x is

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

Answer key · Absolute value and radicals

Math 20-1 · maddyhelps.com

  1. d) 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.
  2. d) 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.
  3. b) 4 — Work inside the bars first: −7 + 3 = −4. Absolute value is distance from zero, so |−4| = 4. The bars are not a minus sign and they are not brackets you can distribute over — they act on whatever the expression inside comes out to.
  4. b) 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.
  5. a) x = 7 or x = −2 — Two expressions have absolute value 9: nine and negative nine. So 2x − 5 = 9 gives x = 7, and 2x − 5 = −9 gives x = −2. Both check. Solving only the positive case is the usual way to lose half the answer.
  6. b) 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.
  7. c) −x — Absolute value is never negative. When x is already negative, −x is the positive version of it: if x = −6 then −x = 6, which is |−6|. Writing −x looks like a negative answer, but the minus sign is what makes it positive here.
  8. c) 30√2 — Multiply the outside numbers and the radicands separately: 2 × 5 = 10 and √3 · √6 = √18. Then √18 = 3√2, so 10√18 = 30√2. Leaving it as 10√18 is not wrong arithmetic, but it is not simplest form.
  9. d) 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.
  10. c) x = 1 only — Case one: x − 3 = 2x gives x = −3, but then the right side is 2(−3) = −6, and an absolute value can never equal a negative. Reject it. Case two: x − 3 = −2x gives x = 1, and |1 − 3| = 2 = 2(1) ✓. When the other side contains the variable, checking is not optional.