PSAT Math Quiz: Radicals And Absolute Values

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.

Question 1

1. $2xy\sqrt{3y}$ (correct answer)
2. $x\sqrt{12y^3}$
3. $2x\sqrt{3y^3}$
4. $2x^2y\sqrt{3y}$

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

1. $25a$
2. $7a$ (correct answer)
3. $\sqrt{25a^2}$
4. $5a^2$

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 3

1. $6\sqrt{2}$
2. $5\sqrt{2}$ (correct answer)
3. $\sqrt{82}$
4. $10\sqrt{2}$

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 4

1. $\dfrac{3(2-\sqrt{5})}{-1}$
2. $\dfrac{3(2+\sqrt{5})}{-1}$ (correct answer)
3. $\dfrac{6+3\sqrt{5}}{9}$
4. $6+3\sqrt{5}$

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 5

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

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 6

1. $x=0$
2. $x=3$ (correct answer)
3. $x=6$
4. $x=\pm3$

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 7

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

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 $x$ to $a$ is $|x - a|$.

Question 8

1. $x=0$
2. $x=8$ (correct answer)
3. $x=\pm8$
4. 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 9

1. $\dfrac{7}{8}$ (correct answer)
2. $\dfrac{49}{8}$
3. $\dfrac{8}{7}$
4. $\dfrac{7}{64}$

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 10

1. $(-3,7)$
2. $(-\infty,-3]\cup[7,\infty)$ (correct answer)
3. $(-\infty,7]\cup[-3,\infty)$
4. $[-3,7]$

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 11

1. $\dfrac{5\sqrt{3}}{\sqrt{3}}$
2. $5$ (correct answer)
3. $\sqrt{39}$
4. $\dfrac{39}{3}$

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 12

1. $\sqrt{39}$
2. $18$ (correct answer)
3. $6\sqrt{3}$
4. $9\sqrt{2}$

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 13

1. $5x\sqrt{3x}$ (correct answer)
2. $25x\sqrt{3}$
3. $5\sqrt{3x^3}$
4. $15\sqrt{x}$

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 14

1. $x=6$
2. $x=-4$
3. $x=6$ or $-4$ (correct answer)
4. $x=5$ 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 15

1. $x=4$
2. $x=8$
3. $x=\dfrac{17}{2}$ (correct answer)
4. 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 16

1. $7\sqrt{2}+2\sqrt{2}$
2. $9\sqrt{2}$ (correct answer)
3. $\sqrt{106}$
4. $11\sqrt{2}$

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 17

1. $\{-5,13\}$ (correct answer)
2. $\{5,13\}$
3. $\{-13,5\}$
4. $\{-5,9\}$

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 18

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

Question 19

1. $x=1$
2. $x=8$
3. $x=-1$ or $8$ (correct answer)
4. $x=1$ or $8$

Explanation: To solve the absolute value equation $|2x-7| = 9$, we need to consider that the expression inside the absolute value bars can equal either 9 or -9. This gives us two cases: Case 1: $2x - 7 = 9$, which yields $2x = 16$, so $x = 8$. Case 2: $2x - 7 = -9$, which yields $2x = -2$, so $x = -1$. We can verify: when $x = 8$, $|2(8)-7| = |16-7| = |9| = 9$ ✓, and when $x = -1$, $|2(-1)-7| = |-2-7| = |-9| = 9$ ✓. The most common error is solving only one case and forgetting that absolute value represents distance, which can be achieved in two directions. Remember that $|A| = k$ means $A = k$ or $A = -k$ when $k > 0$.

Question 20

1. $|x+2|=5;\ x=3,-7$ (correct answer)
2. $|x-2|=5;\ x=7,-3$
3. $|x+2|=5;\ x=7,-3$
4. $|x-2|=5;\ x=3,-7$

Explanation: The question requires the equation and solutions for a point 5 units from -2 on the number line. The absolute value represents distance, so |x - (-2)| = 5 simplifies to |x + 2| = 5. Solve the two cases: x + 2 = 5 gives x = 3, and x + 2 = -5 gives x = -7. Both satisfy the distance condition. A common error is switching the sign in the equation, like using |x - 2| which would be distance from 2 instead. Verify by calculating distances: from 3 to -2 is 5, from -7 to -2 is 5. When solving absolute values, always consider both positive and negative cases and check solutions.