ACCUPLACER Advanced Algebra & Functions Quiz: Radical Operations
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Radical OperationsQuestion 1 of 20

Which of the following is equivalent to 350−323\sqrt{50} - \sqrt{32}?

3183\sqrt{18}
727\sqrt{2}
11211\sqrt{2}
19219\sqrt{2}
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ACCUPLACER Advanced Algebra & Functions Quiz

ACCUPLACER Advanced Algebra & Functions Quiz: Radical Operations

Practice Radical Operations in ACCUPLACER Advanced Algebra & Functions with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Radical Operations, giving you a quick way to practice the rules, question types, and explanations that matter most for ACCUPLACER Advanced Algebra & Functions.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Which of the following is equivalent to 350−323\sqrt{50} - \sqrt{32}?

  1. 3183\sqrt{18}
  2. 727\sqrt{2}
  3. 11211\sqrt{2} (correct answer)
  4. 19219\sqrt{2}
Explanation: To combine the radicals, they must first be simplified to have the same radicand. 350=325⋅2=3⋅52=1523\sqrt{50} = 3\sqrt{25 \cdot 2} = 3 \cdot 5\sqrt{2} = 15\sqrt{2}. And 32=16⋅2=42\sqrt{32} = \sqrt{16 \cdot 2} = 4\sqrt{2}. The expression becomes 152−4215\sqrt{2} - 4\sqrt{2}, which simplifies to (15−4)2=112(15-4)\sqrt{2} = 11\sqrt{2}.

Question 2

For x>0x > 0, which of the following is equivalent to 48x3+x12x\sqrt{48x^3} + x\sqrt{12x}?

  1. 5x3x5x\sqrt{3x}
  2. 6x3x6x\sqrt{3x} (correct answer)
  3. 8x3x8x\sqrt{3x}
  4. 20x3x20x\sqrt{3x}
Explanation: First, simplify each radical term. For the first term, 48x3=16⋅3⋅x2⋅x=4x3x\sqrt{48x^3} = \sqrt{16 \cdot 3 \cdot x^2 \cdot x} = 4x\sqrt{3x}. For the second term, x12x=x4⋅3x=x⋅23x=2x3xx\sqrt{12x} = x\sqrt{4 \cdot 3x} = x \cdot 2\sqrt{3x} = 2x\sqrt{3x}. Now, add the like terms: 4x3x+2x3x=6x3x4x\sqrt{3x} + 2x\sqrt{3x} = 6x\sqrt{3x}.

Question 3

Which of the following expressions is equivalent to 3−23+2\frac{\sqrt{3} - \sqrt{2}}{\sqrt{3} + \sqrt{2}}?

  1. 1
  2. 5
  3. 5−65 - \sqrt{6}
  4. 5−265 - 2\sqrt{6} (correct answer)
Explanation: To simplify, rationalize the denominator by multiplying the numerator and denominator by the conjugate of the denominator, which is 3−2\sqrt{3} - \sqrt{2}. Numerator: (3−2)(3−2)=(3)2−2(3)(2)+(2)2=3−26+2=5−26(\sqrt{3} - \sqrt{2})(\sqrt{3} - \sqrt{2}) = (\sqrt{3})^2 - 2(\sqrt{3})(\sqrt{2}) + (\sqrt{2})^2 = 3 - 2\sqrt{6} + 2 = 5 - 2\sqrt{6}. Denominator: (3+2)(3−2)=(3)2−(2)2=3−2=1(\sqrt{3} + \sqrt{2})(\sqrt{3} - \sqrt{2}) = (\sqrt{3})^2 - (\sqrt{2})^2 = 3 - 2 = 1. The expression simplifies to 5−261=5−26\frac{5 - 2\sqrt{6}}{1} = 5 - 2\sqrt{6}.

Question 4

If the area of a rectangle is 7 and its width is 3−23 - \sqrt{2}, what is its length?

  1. 21−7221 - 7\sqrt{2}
  2. 3+23 + \sqrt{2} (correct answer)
  3. 3−23 - \sqrt{2}
  4. 1
Explanation: Area of a rectangle is length times width (A = l * w). To find the length, divide the area by the width: l=Aw=73−2l = \frac{A}{w} = \frac{7}{3 - \sqrt{2}}. To simplify, rationalize the denominator by multiplying by the conjugate, 3+23 + \sqrt{2}: 73−2⋅3+23+2=7(3+2)32−(2)2=21+729−2=21+727\frac{7}{3 - \sqrt{2}} \cdot \frac{3 + \sqrt{2}}{3 + \sqrt{2}} = \frac{7(3 + \sqrt{2})}{3^2 - (\sqrt{2})^2} = \frac{21 + 7\sqrt{2}}{9 - 2} = \frac{21 + 7\sqrt{2}}{7}. Dividing each term in the numerator by 7 gives 3+23 + \sqrt{2}.

Question 5

Which of the following values is the greatest?

  1. 2112\sqrt{11}
  2. 353\sqrt{5}
  3. 434\sqrt{3}
  4. 525\sqrt{2} (correct answer)
Explanation: To compare the values, it is helpful to square each one to remove the radical. (211)2=4⋅11=44(2\sqrt{11})^2 = 4 \cdot 11 = 44. (35)2=9⋅5=45(3\sqrt{5})^2 = 9 \cdot 5 = 45. (43)2=16⋅3=48(4\sqrt{3})^2 = 16 \cdot 3 = 48. (52)2=25⋅2=50(5\sqrt{2})^2 = 25 \cdot 2 = 50. Since 50 is the largest squared value, 525\sqrt{2} is the greatest of the original values.

Question 6

Which of the following is equivalent to the expression 633\frac{6}{\sqrt[3]{3}}?

  1. 2332\sqrt[3]{3}
  2. 6336\sqrt[3]{3}
  3. 2932\sqrt[3]{9} (correct answer)
  4. 636\sqrt{3}
Explanation: To rationalize a cube root in the denominator, you need to make the radicand a perfect cube. The radicand is 3, which is 313^1. To make it a perfect cube (333^3), we need to multiply by 32=93^2 = 9. So, we multiply the numerator and denominator by 93\sqrt[3]{9}: 633⋅9393=693273=6933\frac{6}{\sqrt[3]{3}} \cdot \frac{\sqrt[3]{9}}{\sqrt[3]{9}} = \frac{6\sqrt[3]{9}}{\sqrt[3]{27}} = \frac{6\sqrt[3]{9}}{3}. This simplifies to 2932\sqrt[3]{9}.

Question 7

The expression 54+121\sqrt{54 + \sqrt{121}} simplifies to which of the following?

  1. 65\sqrt{65} (correct answer)
  2. 36+113\sqrt{6} + 11
  3. 54+11\sqrt{54} + 11
  4. 14
Explanation: According to the order of operations, simplify the innermost radical first. 121=11\sqrt{121} = 11. The expression becomes 54+11\sqrt{54 + 11}. Adding the numbers inside the radical gives 65\sqrt{65}. Since 65 has no perfect square factors (its factors are 5 and 13), the radical cannot be simplified further.

Question 8

If x=3x = \sqrt{3}, what is the value of the expression x4−9x−3\frac{x^4 - 9}{x - \sqrt{3}}?

  1. 0
  2. 1
  3. 636\sqrt{3}
  4. undefined (correct answer)
Explanation: First, substitute 3\sqrt{3} for x in the denominator. The denominator becomes 3−3=0\sqrt{3} - \sqrt{3} = 0. Since division by zero is undefined, the expression is undefined for x=3x = \sqrt{3}. Although the numerator x4−9=(3)4−9=32−9=9−9=0x^4-9 = (\sqrt{3})^4-9 = 3^2-9=9-9=0, the form 0/0 is indeterminate, but as a direct evaluation, the expression is undefined because of the zero in the denominator.

Question 9

Which of the following is equivalent to 12+1+13+2+12+3\frac{1}{\sqrt{2}+1} + \frac{1}{\sqrt{3}+\sqrt{2}} + \frac{1}{2+\sqrt{3}}?

  1. 1 (correct answer)
  2. -1
  3. 3
  4. 232\sqrt{3}
Explanation: This is a telescoping sum. Rationalize each term separately. 12+1=2−1(2)2−12=2−11=2−1\frac{1}{\sqrt{2}+1} = \frac{\sqrt{2}-1}{(\sqrt{2})^2-1^2} = \frac{\sqrt{2}-1}{1} = \sqrt{2}-1. 13+2=3−2(3)2−(2)2=3−21=3−2\frac{1}{\sqrt{3}+\sqrt{2}} = \frac{\sqrt{3}-\sqrt{2}}{(\sqrt{3})^2-(\sqrt{2})^2} = \frac{\sqrt{3}-\sqrt{2}}{1} = \sqrt{3}-\sqrt{2}. 12+3=2−322−(3)2=2−31=2−3\frac{1}{2+\sqrt{3}} = \frac{2-\sqrt{3}}{2^2-(\sqrt{3})^2} = \frac{2-\sqrt{3}}{1} = 2-\sqrt{3}. Now add the simplified terms: (2−1)+(3−2)+(2−3)(\sqrt{2}-1) + (\sqrt{3}-\sqrt{2}) + (2-\sqrt{3}). The terms 2\sqrt{2} and −2-\sqrt{2} cancel, and the terms 3\sqrt{3} and −3-\sqrt{3} cancel, leaving −1+2=1-1 + 2 = 1.

Question 10

Which of the following is equivalent to 753−48\frac{\sqrt{75}}{3} - \sqrt{48}?

  1. −33-3\sqrt{3}
  2. −733-\frac{7\sqrt{3}}{3} (correct answer)
  3. 533−43\frac{5\sqrt{3}}{3} - 4\sqrt{3}
  4. 27\sqrt{27}
Explanation: First, simplify both radical terms. 75=25⋅3=53\sqrt{75} = \sqrt{25 \cdot 3} = 5\sqrt{3}. So the first term is 533\frac{5\sqrt{3}}{3}. The second term is 48=16⋅3=43\sqrt{48} = \sqrt{16 \cdot 3} = 4\sqrt{3}. The expression is now 533−43\frac{5\sqrt{3}}{3} - 4\sqrt{3}. To subtract, find a common denominator: 533−1233=53−1233=−733\frac{5\sqrt{3}}{3} - \frac{12\sqrt{3}}{3} = \frac{5\sqrt{3} - 12\sqrt{3}}{3} = \frac{-7\sqrt{3}}{3}.

Question 11

Which expression is equivalent to x2+6x+9\sqrt{x^2 + 6x + 9} for x<−3x < -3?

  1. −x−3-x-3 (correct answer)
  2. x+3x+3
  3. x−3x-3
  4. −x+3-x+3
Explanation: The expression under the radical, x2+6x+9x^2 + 6x + 9, is a perfect square trinomial, which factors to (x+3)2(x+3)^2. Therefore, x2+6x+9=(x+3)2\sqrt{x^2 + 6x + 9} = \sqrt{(x+3)^2}. The square root of a squared expression is its absolute value: ∣x+3∣|x+3|. Since the problem states that x<−3x < -3, the value of x+3x+3 will be negative. The absolute value of a negative number is its opposite. Therefore, ∣x+3∣=−(x+3)=−x−3|x+3| = -(x+3) = -x-3.

Question 12

Which of the following is the simplified form of (32−23)2(3\sqrt{2} - 2\sqrt{3})^2?

  1. 6
  2. 30
  3. 30−6630 - 6\sqrt{6}
  4. 30−12630 - 12\sqrt{6} (correct answer)
Explanation: To square the binomial, use the formula (a−b)2=a2−2ab+b2(a-b)^2 = a^2 - 2ab + b^2. Here, a=32a = 3\sqrt{2} and b=23b = 2\sqrt{3}. So, a2=(32)2=9⋅2=18a^2 = (3\sqrt{2})^2 = 9 \cdot 2 = 18. b2=(23)2=4⋅3=12b^2 = (2\sqrt{3})^2 = 4 \cdot 3 = 12. The middle term is −2ab=−2(32)(23)=−126-2ab = -2(3\sqrt{2})(2\sqrt{3}) = -12\sqrt{6}. Combining these gives 18−126+12=30−12618 - 12\sqrt{6} + 12 = 30 - 12\sqrt{6}.

Question 13

Which of the following is equivalent to 403+1353\sqrt[3]{40} + \sqrt[3]{135}?

  1. 1753\sqrt[3]{175}
  2. 5535\sqrt[3]{5} (correct answer)
  3. 115311\sqrt[3]{5}
  4. 253+5332\sqrt[3]{5} + 5\sqrt[3]{3}
Explanation: First, simplify each cube root by finding perfect cube factors. 403=8⋅53=8353=253\sqrt[3]{40} = \sqrt[3]{8 \cdot 5} = \sqrt[3]{8}\sqrt[3]{5} = 2\sqrt[3]{5}. Then, 1353=27⋅53=27353=353\sqrt[3]{135} = \sqrt[3]{27 \cdot 5} = \sqrt[3]{27}\sqrt[3]{5} = 3\sqrt[3]{5}. Now, add the like radicals: 253+353=5532\sqrt[3]{5} + 3\sqrt[3]{5} = 5\sqrt[3]{5}.

Question 14

Which of the following is equivalent to 42−6\frac{4}{2 - \sqrt{6}}?

  1. −4−26-4 - 2\sqrt{6} (correct answer)
  2. −4+26-4 + 2\sqrt{6}
  3. 4+264 + 2\sqrt{6}
  4. 4+265\frac{4+2\sqrt{6}}{5}
Explanation: To rationalize the denominator, multiply the numerator and denominator by the conjugate of the denominator, which is 2+62 + \sqrt{6}. This gives 42−6⋅2+62+6=4(2+6)(2)2−(6)2=8+464−6=8+46−2\frac{4}{2 - \sqrt{6}} \cdot \frac{2 + \sqrt{6}}{2 + \sqrt{6}} = \frac{4(2 + \sqrt{6})}{(2)^2 - (\sqrt{6})^2} = \frac{8 + 4\sqrt{6}}{4 - 6} = \frac{8 + 4\sqrt{6}}{-2}. Dividing each term in the numerator by -2 gives 8−2+46−2=−4−26\frac{8}{-2} + \frac{4\sqrt{6}}{-2} = -4 - 2\sqrt{6}.

Question 15

For x>0x > 0, which expression is equivalent to x3/2(x1/2+2x−1/2)x^{3/2} (x^{1/2} + 2x^{-1/2})?

  1. x2+2x^2 + 2
  2. x2+2xx^2 + 2x (correct answer)
  3. x3/4+2x−3/4x^{3/4} + 2x^{-3/4}
  4. x2+2xx^2 + 2\sqrt{x}
Explanation: Distribute x3/2x^{3/2} to both terms inside the parentheses. The first term becomes x3/2⋅x1/2=x3/2+1/2=x4/2=x2x^{3/2} \cdot x^{1/2} = x^{3/2 + 1/2} = x^{4/2} = x^2. The second term becomes x3/2⋅2x−1/2=2x3/2−1/2=2x2/2=2x1=2xx^{3/2} \cdot 2x^{-1/2} = 2x^{3/2 - 1/2} = 2x^{2/2} = 2x^1 = 2x. The sum is x2+2xx^2 + 2x.

Question 16

The legs of a right triangle have lengths 5\sqrt{5} inches and 232\sqrt{3} inches. What is the perimeter of the triangle?

  1. 17\sqrt{17} inches
  2. 5+23+11\sqrt{5} + 2\sqrt{3} + \sqrt{11} inches
  3. 5+23+17\sqrt{5} + 2\sqrt{3} + \sqrt{17} inches (correct answer)
  4. 34\sqrt{34} inches
Explanation: First, find the length of the hypotenuse, c, using the Pythagorean theorem, a2+b2=c2a^2 + b^2 = c^2. Let a=5a = \sqrt{5} and b=23b = 2\sqrt{3}. Then c2=(5)2+(23)2=5+(4⋅3)=5+12=17c^2 = (\sqrt{5})^2 + (2\sqrt{3})^2 = 5 + (4 \cdot 3) = 5 + 12 = 17. So, c=17c = \sqrt{17}. The perimeter is the sum of the lengths of the three sides: 5+23+17\sqrt{5} + 2\sqrt{3} + \sqrt{17}. These radicals cannot be simplified or combined further.

Question 17

Which of the following is equivalent to 1−1x1+1x\frac{1 - \frac{1}{\sqrt{x}}}{1 + \frac{1}{\sqrt{x}}} for x>0,x≠1x>0, x \neq 1?

  1. x−1x+1\frac{\sqrt{x}-1}{\sqrt{x}+1}
  2. x−1x+1\frac{x-1}{x+1}
  3. x−2x+1x−1\frac{x - 2\sqrt{x} + 1}{x-1} (correct answer)
  4. -1
Explanation: First, simplify the complex fraction by multiplying the numerator and the denominator by x\sqrt{x}. This yields x(1−1x)x(1+1x)=x−1x+1\frac{\sqrt{x}(1 - \frac{1}{\sqrt{x}})}{\sqrt{x}(1 + \frac{1}{\sqrt{x}})} = \frac{\sqrt{x} - 1}{\sqrt{x} + 1}. To simplify further, we must rationalize the denominator by multiplying by its conjugate, x−1\sqrt{x} - 1. This gives (x−1)(x−1)(x+1)(x−1)=(x)2−2x+1(x)2−12=x−2x+1x−1\frac{(\sqrt{x} - 1)(\sqrt{x} - 1)}{(\sqrt{x} + 1)(\sqrt{x} - 1)} = \frac{(\sqrt{x})^2 - 2\sqrt{x} + 1}{(\sqrt{x})^2 - 1^2} = \frac{x - 2\sqrt{x} + 1}{x-1}.

Question 18

For x>0x > 0 and y>0y > 0, what is the product of 18xy2\sqrt{18xy^2} and 6x3y\sqrt{6x^3y} in simplest form?

  1. 3x2y12y3x^2y\sqrt{12y}
  2. 6x23y36x^2\sqrt{3y^3}
  3. 6x2y3y6x^2y\sqrt{3y} (correct answer)
  4. 6x2y36x^2y\sqrt{3}
Explanation: First, multiply the expressions under the radical: (18xy2)(6x3y)=108x4y3\sqrt{(18xy^2)(6x^3y)} = \sqrt{108x^4y^3}. Now, simplify the resulting radical. The number 108 can be factored as 36⋅336 \cdot 3. The variable parts are x4x^4 and y3=y2⋅yy^3 = y^2 \cdot y. So, 36⋅3⋅x4⋅y2⋅y=36x4y23y=6x2y3y\sqrt{36 \cdot 3 \cdot x^4 \cdot y^2 \cdot y} = \sqrt{36}\sqrt{x^4}\sqrt{y^2}\sqrt{3y} = 6x^2y\sqrt{3y}.

Question 19

Which of the following is equivalent to (23−5)(3+35)(2\sqrt{3} - \sqrt{5})(\sqrt{3} + 3\sqrt{5})?

  1. −9-9
  2. −9+515-9 + 5\sqrt{15} (correct answer)
  3. −9+715-9 + 7\sqrt{15}
  4. 21+51521 + 5\sqrt{15}
Explanation: Use the FOIL method to multiply the binomials. First: (23)(3)=2⋅3=6(2\sqrt{3})(\sqrt{3}) = 2 \cdot 3 = 6. Outer: (23)(35)=615(2\sqrt{3})(3\sqrt{5}) = 6\sqrt{15}. Inner: (−5)(3)=−15(-\sqrt{5})(\sqrt{3}) = -\sqrt{15}. Last: (−5)(35)=−3⋅5=−15(-\sqrt{5})(3\sqrt{5}) = -3 \cdot 5 = -15. Combine the terms: 6+615−15−15=(6−15)+(615−15)=−9+5156 + 6\sqrt{15} - \sqrt{15} - 15 = (6 - 15) + (6\sqrt{15} - \sqrt{15}) = -9 + 5\sqrt{15}.

Question 20

Which of the following integers is the value of (50−2)2(\sqrt{50} - \sqrt{2})^2?

  1. 24
  2. 32 (correct answer)
  3. 48
  4. 52
Explanation: First, simplify the expression inside the parentheses. 50=25⋅2=52\sqrt{50} = \sqrt{25 \cdot 2} = 5\sqrt{2}. The expression becomes (52−2)2(5\sqrt{2} - \sqrt{2})^2. Combining the like terms gives (42)2(4\sqrt{2})^2. Squaring this expression results in 42⋅(2)2=16⋅2=324^2 \cdot (\sqrt{2})^2 = 16 \cdot 2 = 32.