PSAT Math Quiz: Radicals And Absolute Values

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Radicals And Absolute ValuesQuestion 1 of 20

Which expression is equivalent to $\sqrt{12x^2y^3}$ for $x\ge 0$ and $y\ge 0$ ? Make sure your answer has no perfect-square factors left under the radical; a common error is forgetting that $\sqrt{x^2}=x$ only because $x\ge 0$ is given.

2xy\sqrt{3y}

x\sqrt{12y^3}

2x\sqrt{3y^3}

2x^2y\sqrt{3y}

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PSAT Math Quiz

PSAT Math Quiz: Radicals And Absolute Values

Practice Radicals And Absolute Values in PSAT Math 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 Radicals And Absolute Values, giving you a quick way to practice the rules, question types, and explanations that matter most for PSAT Math.

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

$2xy\sqrt{3y}$ (correct answer)
$x\sqrt{12y^3}$
$2x\sqrt{3y^3}$
$2x^2y\sqrt{3y}$

Explanation: We need to simplify

\sqrt{12x^2y^3}

where

x \geq 0

and

y \geq 0

. First, factor the expression under the radical to identify perfect squares:

12x^2y^3 = 4 \cdot 3 \cdot x^2 \cdot y^2 \cdot y = 4x^2y^2 \cdot 3y

. Now we can take the square root of the perfect square factors:

\sqrt{4x^2y^2 \cdot 3y} = \sqrt{4x^2y^2} \cdot \sqrt{3y} = 2xy\sqrt{3y}

. Note that

\sqrt{x^2} = x

(not

|x|

) because we're given that

x \geq 0

. A common error is leaving perfect square factors under the radical or incorrectly handling the variable exponents. Always factor completely and extract all perfect squares when simplifying radicals.

Question 2

Which expression equals $\dfrac{5}{2+\sqrt3}$ when the denominator is rationalized?

$5+\sqrt3$
$10-5\sqrt3$ (correct answer)
$10+5\sqrt3$
$5-\sqrt3$

Explanation: When you encounter a fraction with a radical in the denominator, you need to rationalize it by eliminating the square root from the bottom. This process uses the conjugate method. To rationalize

\dfrac{5}{2+\sqrt{3}}

, multiply both numerator and denominator by the conjugate of the denominator. The conjugate of

2+\sqrt{3}

2-\sqrt{3}

\dfrac{5}{2+\sqrt{3}} \cdot \dfrac{2-\sqrt{3}}{2-\sqrt{3}} = \dfrac{5(2-\sqrt{3})}{(2+\sqrt{3})(2-\sqrt{3})}

The numerator becomes:

5(2-\sqrt{3}) = 10-5\sqrt{3}

The denominator uses the difference of squares pattern:

(2+\sqrt{3})(2-\sqrt{3}) = 2^2 - (\sqrt{3})^2 = 4-3 = 1

Therefore:

\dfrac{10-5\sqrt{3}}{1} = 10-5\sqrt{3}

, which is answer choice B. Looking at the wrong answers: A)

5+\sqrt{3}

might result from incorrectly adding terms or confusing the rationalization process. C)

10+5\sqrt{3}

comes from using the wrong sign—multiplying by

2+\sqrt{3}

instead of the conjugate

2-\sqrt{3}

. D)

5-\sqrt{3}

suggests someone forgot to distribute the 5 properly in the numerator. Study tip: Always use the conjugate when rationalizing denominators containing addition or subtraction with radicals. The conjugate flips the middle sign, and when multiplied together, the radicals eliminate due to the difference of squares pattern:

(a+b)(a-b) = a^2-b^2

Question 3

Which expression is equivalent to $\sqrt{9a^2}+\sqrt{16a^2}$ for $a\ge0$ ?

$25a$
$7a$ (correct answer)
$\sqrt{25a^2}$
$5a^2$

Explanation: The question asks for the expression equivalent to √(

9a^2

) + √(

16a^2

) for a ≥ 0. Simplify each: √(

9a^2

) = 3a, √(

16a^2

) = 4a, so 3a + 4a = 7a. This assumes a ≥ 0 to avoid absolute values. Options like 25a might come from multiplying instead of adding. A key error is treating them as √(

9a^2

16a^2

), which is incorrect. Verify: for a=1, √9 + √16 = 3+4=7, matches 7a. When adding simplified radicals, handle each separately before combining.

Question 4

What is the sum of all real solutions to $\sqrt{3x-2}=|x-2|\;?$

$5$
$7$ (correct answer)
$8$
$13$

Explanation: When you encounter an equation with both a square root and absolute value, you need to consider the domain restrictions and cases where the absolute value expression changes sign. First, note that

\sqrt{3x-2}

requires

3x-2 \geq 0

, so

x \geq \frac{2}{3}

. Next, since

|x-2|

equals

x-2

when

x \geq 2

and equals

-(x-2) = 2-x

when

x < 2

, you need to solve two cases. Case 1:

\frac{2}{3} \leq x < 2

Here

|x-2| = 2-x

, so the equation becomes

\sqrt{3x-2} = 2-x

. Squaring both sides:

3x-2 = (2-x)^2 = 4-4x+x^2

. Rearranging:

x^2-7x+6 = 0

, which factors as

(x-1)(x-6) = 0

. This gives

x = 1

x = 6

. Since we need

\frac{2}{3} \leq x < 2

, only

x = 1

works in this case. Case 2:

x \geq 2

Here

|x-2| = x-2

, so

\sqrt{3x-2} = x-2

. Squaring:

3x-2 = (x-2)^2 = x^2-4x+4

. Rearranging:

x^2-7x+6 = 0

. Again we get

x = 1

x = 6

. Since we need

x \geq 2

, only

x = 6

works here. Verify both solutions:

\sqrt{3(1)-2} = 1

and

|1-2| = 1

✓;

\sqrt{3(6)-2} = 4

and

|6-2| = 4

✓. The sum is

1 + 6 = 7

, which is choice B. Choices A, C, and D represent incorrect sums that might arise from including extraneous solutions or computational errors when solving the quadratic. Strategy tip: Always check your domain restrictions first, then verify solutions by substituting back into the original equation—not the squared version.

Question 5

Simplify the expression $\sqrt{72}-2\sqrt{8}+\sqrt{18}$ . Be careful to rewrite each radical using a perfect-square factor before combining like terms; one common mistake is to combine terms that are not like radicals or to stop at an unsimplified radical.

$6\sqrt{2}$
$5\sqrt{2}$ (correct answer)
$\sqrt{82}$
$10\sqrt{2}$

Explanation: This problem asks us to simplify an expression with three radical terms by combining like radicals. First, we need to simplify each radical by factoring out perfect squares:

\sqrt{72} = \sqrt{36 \cdot 2} = 6\sqrt{2}

\sqrt{8} = \sqrt{4 \cdot 2} = 2\sqrt{2}

, and

\sqrt{18} = \sqrt{9 \cdot 2} = 3\sqrt{2}

. Now we can substitute these simplified forms:

6\sqrt{2} - 2(2\sqrt{2}) + 3\sqrt{2} = 6\sqrt{2} - 4\sqrt{2} + 3\sqrt{2}

. Since all terms contain

\sqrt{2}

, we can combine the coefficients:

(6 - 4 + 3)\sqrt{2} = 5\sqrt{2}

. A common error is trying to combine radicals before simplifying them or incorrectly factoring the numbers under the radicals. When working with radicals on the PSAT, always simplify each radical first before attempting to combine terms.

Question 6

Which expression is equivalent to $\dfrac{3}{2-\sqrt{5}}$ after rationalizing the denominator?

$\dfrac{3(2-\sqrt{5})}{-1}$
$\dfrac{3(2+\sqrt{5})}{-1}$ (correct answer)
$\dfrac{6+3\sqrt{5}}{9}$
$6+3\sqrt{5}$

Explanation: The question requires finding the equivalent expression to 3/(2 - √5) after rationalizing the denominator. Multiply numerator and denominator by the conjugate 2 + √5: 3(2 + √5) / ((2 - √5)(2 + √5)) = 3(2 + √5) / (4 - 5) = 3(2 + √5) / (-1). This simplifies to -3(2 + √5), matching the form in choice B. Verify numerically: original ≈ 3 / (-0.236) ≈ -12.71, and -3(4.236) ≈ -12.71. A key error is using the wrong conjugate or mishandling the negative denominator. Another mistake could be not distributing the negative sign correctly. When rationalizing, always check your work by verifying numerical equivalence.

Question 7

A student claims $\sqrt{a}+\sqrt{b}=\sqrt{a+b}$ . Which choice gives a counterexample using specific numbers?

$a=1,b=4$ (correct answer)
$a=0,b=9$
$a=9,b=0$
$a=0,b=0$

Explanation: The question asks for a counterexample to the claim that (

\sqrt{a}

\sqrt{b}

\sqrt{a + b}

) using specific numbers. Test choice A with a=1, b=4: left side is (

\sqrt{1}

\sqrt{4}

= 1 + 2 = 3), right side is (

\sqrt{5}

\approx

2.236), which are not equal, so it's a counterexample. Choices like B (a=0, b=9) give both sides equal to 3, failing as a counterexample since the claim holds. A common error is assuming the claim is true when one variable is zero, but that doesn't disprove it. Another mistake is not calculating both sides fully to compare. When disproving statements, try simple positive numbers and always compute both sides to verify inequality.

Question 8

Solve the equation $|x|=|x-6|$ . Interpret absolute value as distance on the number line.

$x=0$
$x=3$ (correct answer)
$x=6$
$x=\pm3$

Explanation: The question asks to solve |x| = |x - 6| interpreting as distances on the number line. This means distance from x to 0 equals distance to 6, so x is midpoint: x = 3. Check: |3| = 3, |3-6| = 3, equal. Algebraically, consider cases: if x ≥ 6, x = x-6 impossible; if 0 ≤ x < 6, x = 6 - x, 2x=6, x=3; if x < 0, -x = 6 - x impossible. Only x=3. A key error is assuming symmetric solutions like ±3. Always check all intervals defined by critical points.

Question 9

A point on a number line is at position $x$ . Its distance from $-3$ is 5 units. Which equation represents this situation, and what are the solutions? Interpreting absolute value as distance helps avoid sign errors.

$|x+3|=5;\ x=2,-8$ (correct answer)
$|x-3|=5;\ x=8,-2$
$|x+3|=5;\ x=-2,8$
$|x-3|=5;\ x=2,-8$

Explanation: This problem asks us to translate a distance statement into an absolute value equation. If a point at position

x

is 5 units away from -3, then the distance between

x

and -3 equals 5. The distance formula on a number line is

|x - (-3)| = |x + 3| = 5

. To solve this, we consider two cases:

x + 3 = 5

, giving

x = 2

, and

x + 3 = -5

, giving

x = -8

. We can verify: the distance from 2 to -3 is

|2-(-3)| = |5| = 5

✓, and the distance from -8 to -3 is

|-8-(-3)| = |-5| = 5

✓. A common error is writing the equation as

|x - 3| = 5

, which would find points 5 units from positive 3 instead. When setting up distance problems, remember that distance from

x

a

|x - a|

Question 10

Solve the equation $|x-8|=0$ . What is the value of $x$ ?

$x=0$
$x=8$ (correct answer)
$x=\pm8$
No solution

Explanation: The question asks to solve the equation (|x - 8| = 0) and find the value of x. The absolute value equals zero only when the expression inside is zero, so (x - 8 = 0), which gives (x = 8). There are no two cases here since it's exactly zero, not an inequality. A common error is treating it like (|x| = 8) and getting (

\pm

8), but that's for a positive value. Another mistake could be thinking there's no solution, but absolute value is always nonnegative. When solving absolute value equations, emphasize the definition and check if the right side is zero, positive, or negative for validity.

Question 11

Simplify $\sqrt{\dfrac{49}{64}}$ and give the exact value.

$\dfrac{7}{8}$ (correct answer)
$\dfrac{49}{8}$
$\dfrac{8}{7}$
$\dfrac{7}{64}$

Explanation: The question requires simplifying √(49/64) to its exact value. Start by recognizing that the square root of a fraction is the square root of the numerator over the square root of the denominator, so √(49/64) = √49 / √64. Simplify √49 = 7 and √64 = 8, yielding 7/8. This is already in simplest form as both are integers with no common factors. A key error might be squaring instead of taking roots or miscalculating √64 as 4 instead of 8. Always verify by squaring back: (7/8)^2 = 49/64, which matches. For radical simplifications, break down into perfect squares to ensure accuracy.

Question 12

Solve the inequality $|x-2|\ge5$ . Give the solution as a union of intervals.

$(-3,7)$
$(-\infty,-3]\cup[7,\infty)$ (correct answer)
$(-\infty,7]\cup[-3,\infty)$
$[-3,7]$

Explanation: The question requires solving the inequality (|x - 2| \geq 5) and giving the solution as a union of intervals. Consider the two cases for absolute value: x - 2 ≥ 5 or x - 2 ≤ -5, so x ≥ 7 or x ≤ -3. The solution is (-∞, -3] ∪ [7, ∞). Verify boundary points: at x=-3, | -3-2 | =5 ≥5; at x=7, |7-2|=5 ≥5. A key error is reversing the inequalities or forgetting the union, leading to intervals like [-3,7]. Another mistake might be using strict inequalities instead of inclusive. For absolute value inequalities, always handle the two cases separately and test points in each interval to confirm.

Question 13

Simplify $\dfrac{\sqrt{12}+\sqrt{27}}{\sqrt{3}}$ in simplest form.

$\dfrac{5\sqrt{3}}{\sqrt{3}}$
$5$ (correct answer)
$\sqrt{39}$
$\dfrac{39}{3}$

Explanation: The question requires simplifying (√12 + √27)/√3. Simplify numerator: √12 = 2√3, √27 = 3√3, sum 5√3. Divide: 5√3 / √3 = 5. This is fully simplified. A common error is not simplifying radicals before dividing. Check: original ≈ (3.464 + 5.196)/1.732 ≈8.66/1.732≈5, matches. Always simplify components before operations with radicals.

Question 14

Simplify the product $\sqrt{27}\,\sqrt{12}$ . A tempting but incorrect approach is to add inside the radicals; instead, multiply first (or simplify each radical) and then write the result in simplest radical form.

$\sqrt{39}$
$18$ (correct answer)
$6\sqrt{3}$
$9\sqrt{2}$

Explanation: To simplify the product

\sqrt{27} \cdot \sqrt{12}

, we can either multiply first then simplify, or simplify each radical first. Using the multiplication property:

\sqrt{27} \cdot \sqrt{12} = \sqrt{27 \cdot 12} = \sqrt{324}

. Since

324 = 18^2

, we have

\sqrt{324} = 18

. Alternatively, simplifying first:

\sqrt{27} = \sqrt{9 \cdot 3} = 3\sqrt{3}

and

\sqrt{12} = \sqrt{4 \cdot 3} = 2\sqrt{3}

, so

(3\sqrt{3})(2\sqrt{3}) = 6 \cdot 3 = 18

. A common error is trying to add the numbers under the radicals instead of multiplying them. Remember that

\sqrt{a} \cdot \sqrt{b} = \sqrt{ab}

, not

\sqrt{a+b}

Question 15

Simplify $\sqrt{75x^3}$ assuming $x\ge0$ . Write the result in simplest radical form.

$5x\sqrt{3x}$ (correct answer)
$25x\sqrt{3}$
$5\sqrt{3x^3}$
$15\sqrt{x}$

Explanation: The question requires simplifying √(

75x^3

) in simplest radical form assuming x ≥ 0. Factor 75x^3 = 25 * 3 * x^2 * x = 25x^2 * (3x). Take square root: √(

25x^2

* 3x) = √(

25x^2

) * √(3x) = 5x √(3x). This is simplest as 3x has no perfect squares. A common error is incomplete factoring, like missing x^2. Check by squaring: (5x √(3x))^2 = 25x^2 * 3x = 75x^3, correct. Always factor out perfect squares when simplifying radicals with variables.

Question 16

Solve the absolute value equation $|x-1|+2=7$ .

$x=6$
$x=-4$
$x=6$ or $-4$ (correct answer)
$x=5$ or $-5$

Explanation: The question requires solving the absolute value equation |x - 1| + 2 = 7. Isolate the absolute value: |x - 1| = 5. Consider two cases: x - 1 = 5 gives x = 6, and x - 1 = -5 gives x = -4. Both satisfy the original: |6-1| + 2 = 7, |-4-1| + 2 = 7. A common error is forgetting to isolate before splitting cases. Always check solutions in the original equation to confirm. For absolute value equations, isolate first then handle positive and negative scenarios.

Question 17

Solve the equation $\sqrt{2x-1}+1=5$ . Be careful not to divide the radical incorrectly; isolate the radical first, then square both sides, and confirm the solution satisfies the original equation.

$x=4$
$x=8$
$x=\dfrac{17}{2}$ (correct answer)
No real solution

Explanation: To solve

\sqrt{2x-1} + 1 = 5

, we first isolate the radical by subtracting 1 from both sides:

\sqrt{2x-1} = 4

. Now we square both sides to eliminate the square root:

(\sqrt{2x-1})^2 = 4^2

, which gives us

2x - 1 = 16

. Solving for

x

2x = 17

, so

x = \frac{17}{2}

. Let's verify by substituting back:

\sqrt{2(\frac{17}{2})-1} + 1 = \sqrt{17-1} + 1 = \sqrt{16} + 1 = 4 + 1 = 5

✓. A common error is trying to distribute operations across the radical before isolating it, such as incorrectly thinking

\sqrt{2x-1} + 1 = \sqrt{2x} + \sqrt{0}

. Always isolate the radical term before squaring both sides.

Question 18

Simplify $\sqrt{98}+\sqrt{8}$ in simplest radical form.

$7\sqrt{2}+2\sqrt{2}$
$9\sqrt{2}$ (correct answer)
$\sqrt{106}$
$11\sqrt{2}$

Explanation: The question requires simplifying √98 + √8 in simplest radical form. Factor: √98 = √(492) = 7√2, √8 = √(42) = 2√2. Add: 7√2 + 2√2 = 9√2. This is simplest as like terms are combined. A key error is adding under one radical, like √(98+8), which is wrong. Check by approximating: 7√2 ≈9.899, 2√2≈2.828, sum≈12.727; 9√2≈12.727, matches. Always simplify each radical before combining.

Question 19

What is the solution set of the absolute value equation $|x-4|=9$ ? (Give both solutions.)

$\{-5,13\}$ (correct answer)
$\{5,13\}$
$\{-13,5\}$
$\{-5,9\}$

Explanation: The question asks for the solution set of the absolute value equation |x - 4| = 9, requiring both solutions. To solve, consider the two cases: x - 4 = 9 or x - 4 = -9, leading to x = 13 or x = -5. The solution set is {-5, 13}. Verify by substitution: | -5 - 4 | = | -9 | = 9, and |13 - 4| = 9, both correct. A common error is only considering the positive case and missing the negative solution. Another mistake might be incorrect arithmetic when solving the linear equations. When solving absolute value equations, always handle both cases and verify solutions to ensure accuracy.

Question 20

Solve the equation $|2x+3|=|x-1|$ . Be careful to consider all sign cases.

$x=-4$ or $-\dfrac{2}{3}$ (correct answer)
$x=4$ or $\dfrac{2}{3}$
$x=-4$ only
$x=-\dfrac{2}{3}$ only

Explanation: The question requires solving |2x+3| = |x-1| considering all sign cases. Square both sides: (2x+3)^2 = (x-1)^2, 4x^2 +12x+9 = x^2 -2x+1, 3x^2 +14x +8=0. Solutions: x = [-14 ± √(196-96)]/6 = [-14 ±10]/6, so x=-4 or x=-2/3. Check: for x=-4, | -8+3|=5, | -4-1|=5; for x=-2/3, | -4/3+3|=5/3, | -2/3-1|=5/3. Both work. A common error is missing a solution by not squaring properly. Always verify in original after solving.