SAT Math Quiz: Radicals And Absolute Values
20 questions · exam conditions
0:00
Radicals And Absolute ValuesQuestion 1 of 20

A point on a number line is at position xx. If its distance from −3-3 is 7 units, which equation correctly models this situation using absolute value?

∣x+3∣=7|x+3|=7
∣x−3∣=7|x-3|=7
∣x+7∣=3|x+7|=3
∣x−7∣=3|x-7|=3
← Back to quizzes

SAT Math Quiz

SAT Math Quiz: Radicals And Absolute Values

Practice Radicals And Absolute Values in SAT 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 SAT 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

A point on a number line is at position xx. If its distance from −3-3 is 7 units, which equation correctly models this situation using absolute value?

  1. ∣x+3∣=7|x+3|=7 (correct answer)
  2. ∣x−3∣=7|x-3|=7
  3. ∣x+7∣=3|x+7|=3
  4. ∣x−7∣=3|x-7|=3
Explanation: This problem asks for an equation modeling the distance between a point at position x and the point -3, where this distance equals 7 units. The distance between x and -3 is given by |x - (-3)| = |x + 3|. Since this distance equals 7, we have |x + 3| = 7. This correctly models the situation because absolute value represents distance on a number line. A common error is confusing the signs and writing |x - 3| = 7, which would represent the distance from positive 3 instead of -3. When setting up distance problems with absolute value, remember that |x - a| represents the distance from x to a.

Question 2

Solve the equation x+5=x−1\sqrt{x+5}=x-1. Because the square root represents a nonnegative value, not every solution to the squared equation will work. Find the real solution(s) and choose the correct answer.​

  1. No real solution
  2. x=4x=4 only (correct answer)
  3. x=0x=0 only
  4. x=0x=0 or 44
Explanation: We need to solve x+5=x−1\sqrt{x+5} = x-1 and check which solutions are valid. Squaring both sides gives x+5=(x−1)2=x2−2x+1x+5 = (x-1)^2 = x^2 - 2x + 1, which simplifies to x2−3x−4=0x^2 - 3x - 4 = 0. Factoring yields (x−4)(x+1)=0(x-4)(x+1) = 0, so x=4x = 4 or x=−1x = -1. However, we must check both solutions: for x=4x = 4, we get 9=3\sqrt{9} = 3 and 4−1=34-1 = 3 ✓; for x=−1x = -1, we get 4=2\sqrt{4} = 2 but −1−1=−2-1-1 = -2 ✗. Since square roots are non-negative, only x=4x = 4 works. Always check solutions when solving radical equations, as squaring can introduce extraneous solutions.

Question 3

Solve the equation x+5=x−1\sqrt{x+5}=x-1. Because the square root represents a nonnegative value, not every solution to the squared equation will work. Find the real solution(s) and choose the correct answer.

  1. x=4x=4 only (correct answer)
  2. No real solution
  3. x=0x=0 or 44
  4. x=0x=0 only
Explanation: We need to solve x+5=x−1\sqrt{x+5} = x-1 and check which solutions are valid. Squaring both sides gives x+5=(x−1)2=x2−2x+1x+5 = (x-1)^2 = x^2 - 2x + 1, which simplifies to x2−3x−4=0x^2 - 3x - 4 = 0. Factoring yields (x−4)(x+1)=0(x-4)(x+1) = 0, so x=4x = 4 or x=−1x = -1. However, we must check both solutions: for x=4x = 4, we get 9=3\sqrt{9} = 3 and 4−1=34-1 = 3 ✓; for x=−1x = -1, we get 4=2\sqrt{4} = 2 but −1−1=−2-1-1 = -2 ✗. Since square roots are non-negative, only x=4x = 4 works. Always check solutions when solving radical equations, as squaring can introduce extraneous solutions.

Question 4

What are all solutions to the equation ∣2x−7∣=5|2x-7|=5?

  1. x=1,6x=1,6 (correct answer)
  2. x=1x=1
  3. x=6x=6
  4. x=−1,6x=-1,6
Explanation: To solve ∣2x−7∣=5|2x-7|=5, we consider two cases based on the definition of absolute value. Case 1: 2x−7=52x-7=5 gives 2x=122x=12, so x=6x=6. Case 2: 2x−7=−52x-7=-5 gives 2x=22x=2, so x=1x=1. Both solutions are valid since substituting back gives ∣2(6)−7∣=∣5∣=5|2(6)-7|=|5|=5 ✓ and ∣2(1)−7∣=∣−5∣=5|2(1)-7|=|-5|=5 ✓. The key insight is that ∣A∣=k|A|=k means either A=kA=k or A=−kA=-k when k>0k>0. Always solve both cases when dealing with absolute value equations.

Question 5

Solve x+9−x=3\sqrt{x+9}-\sqrt{x}=3.

  1. x=0x=0 (correct answer)
  2. x=4x=4
  3. x=9x=9
  4. No solution
Explanation: To solve x+9−x=3\sqrt{x+9}-\sqrt{x}=3, we isolate one radical: x+9=3+x\sqrt{x+9} = 3 + \sqrt{x}. Squaring both sides gives x+9=(3+x)2=9+6x+xx+9 = (3+\sqrt{x})^2 = 9 + 6\sqrt{x} + x. This simplifies to x+9=9+6x+xx+9 = 9 + 6\sqrt{x} + x, or 0=6x0 = 6\sqrt{x}, which means x=0\sqrt{x} = 0, so x=0x = 0. Let's verify: 0+9−0=3−0=3\sqrt{0+9}-\sqrt{0} = 3-0 = 3 ✓. The key is recognizing that when squaring (a+b)2(a+b)^2, we get a2+2ab+b2a^2+2ab+b^2, not just a2+b2a^2+b^2. Always check solutions in radical equations as squaring can introduce extraneous solutions.

Question 6

Which expression is equivalent to 53\dfrac{5}{\sqrt{3}}? Choose the form with a rational denominator; a common incorrect path is multiplying only the denominator by 3\sqrt{3}.

  1. 533\dfrac{5\sqrt{3}}{3} (correct answer)
  2. 533\dfrac{5}{3\sqrt{3}}
  3. 153\dfrac{15}{\sqrt{3}}
  4. 153\dfrac{\sqrt{15}}{3}
Explanation: We need to rationalize the denominator of 53\frac{5}{\sqrt{3}}. To eliminate the radical from the denominator, we multiply both numerator and denominator by 3\sqrt{3}: 53⋅33=533\frac{5}{\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}} = \frac{5\sqrt{3}}{3}. The denominator becomes 3⋅3=3\sqrt{3} \cdot \sqrt{3} = 3. A common mistake is multiplying only the denominator by 3\sqrt{3}, which would incorrectly give 53\frac{5}{3}. When rationalizing, always multiply both parts of the fraction by the same expression.

Question 7

A point PP on a number line is 7 units from −2-2. If xx is the coordinate of PP, which equation represents this situation using absolute value (distance)?

  1. ∣x+2∣=7|x+2|=7 (correct answer)
  2. ∣x−2∣=7|x-2|=7
  3. ∣x+7∣=2|x+7|=2
  4. ∣x−7∣=−2|x-7|=-2
Explanation: This problem asks us to write an equation representing that point P is 7 units from -2 on a number line. The distance between P (with coordinate x) and -2 is given by ∣x−(−2)∣=∣x+2∣|x - (-2)| = |x + 2|. Since this distance equals 7, we have ∣x+2∣=7|x + 2| = 7. This makes sense because solving gives us x+2=7x + 2 = 7 or x+2=−7x + 2 = -7, yielding x=5x = 5 or x=−9x = -9, both of which are indeed 7 units from -2. A common error is confusing the signs and writing ∣x−2∣|x - 2| instead. Remember that distance from point a to point b is ∣b−a∣|b - a|.

Question 8

Solve the absolute value equation ∣3x−7∣=11|3x-7|=11.

  1. x=6x=6 only
  2. x=−43x=-\dfrac{4}{3} only
  3. x=6x=6 or −43-\dfrac{4}{3} (correct answer)
  4. x=43x=\dfrac{4}{3} or 66
Explanation: To solve ∣3x−7∣=11|3x-7| = 11, we must consider two cases since absolute value represents distance from zero. Case 1: 3x−7=113x-7 = 11 gives 3x=183x = 18, so x=6x = 6. Case 2: 3x−7=−113x-7 = -11 gives 3x=−43x = -4, so x=−43x = -\frac{4}{3}. We can verify: ∣3(6)−7∣=∣18−7∣=∣11∣=11|3(6)-7| = |18-7| = |11| = 11 ✓ and ∣3(−43)−7∣=∣−4−7∣=∣−11∣=11|3(-\frac{4}{3})-7| = |-4-7| = |-11| = 11 ✓. The common mistake is solving only the positive case or making sign errors when setting up the negative case. For absolute value equations, always solve both cases: expression = positive value AND expression = negative value.

Question 9

Solve the absolute value equation ∣3x−5∣=7|3x-5|=7.

  1. x=4x=4
  2. x=−23x=-\tfrac{2}{3}
  3. x=4x=4 or −23-\tfrac{2}{3} (correct answer)
  4. x=23x=\tfrac{2}{3} or 44
Explanation: The problem is to solve the absolute value equation |3x - 5| = 7. The absolute value equation |A| = 7 means A = 7 or A = -7, so consider two cases: 3x - 5 = 7 or 3x - 5 = -7. Solving the first: 3x = 12, x = 4; the second: 3x = -2, x = -2/3. Both solutions satisfy the original equation upon checking. A common mistake is solving only one case, such as assuming absolute value always makes it positive and missing the negative branch. Another error could be incorrect arithmetic in solving the linear equations. For absolute value equations, always solve both cases and verify solutions to ensure they work.

Question 10

Simplify the expression 4916\sqrt{\dfrac{49}{16}}. Give an exact value.

  1. 4916\dfrac{49}{16}
  2. −74-\dfrac{7}{4}
  3. 74\dfrac{7}{4} (correct answer)
  4. 47\dfrac{4}{7}
Explanation: To simplify 4916\sqrt{\frac{49}{16}}, we use the property that ab=ab\sqrt{\frac{a}{b}} = \frac{\sqrt{a}}{\sqrt{b}} for positive values. This gives us 4916=74\frac{\sqrt{49}}{\sqrt{16}} = \frac{7}{4}. Since both 49 and 16 are perfect squares, the result is a rational number. The principal square root is always non-negative, so the answer is positive 74\frac{7}{4}. A common error would be to include a negative value, but by definition, x\sqrt{x} represents only the non-negative square root. When simplifying radicals of fractions, apply the square root to both numerator and denominator separately.

Question 11

Solve the absolute value equation ∣2x−7∣=9|2x-7|=9. Give all real solutions.

  1. x=8x=8
  2. x=−1, 8x=-1,\,8 (correct answer)
  3. x=−8, 1x=-8,\,1
  4. x=−1x=-1
Explanation: To solve the absolute value equation ∣2x−7∣=9|2x-7| = 9, we must consider both cases where the expression inside equals 9 or -9. Case 1: 2x−7=92x-7 = 9 gives us 2x=162x = 16, so x=8x = 8. Case 2: 2x−7=−92x-7 = -9 gives us 2x=−22x = -2, so x=−1x = -1. We can verify: when x=8x = 8, ∣2(8)−7∣=∣16−7∣=∣9∣=9|2(8)-7| = |16-7| = |9| = 9 ✓, and when x=−1x = -1, ∣2(−1)−7∣=∣−2−7∣=∣−9∣=9|2(-1)-7| = |-2-7| = |-9| = 9 ✓. The most common error is solving only the positive case and missing half the solutions. Remember that absolute value equations typically have two solutions unless the expression equals zero.

Question 12

Simplify the expression 72−218+8\sqrt{72}-2\sqrt{18}+\sqrt{8}.

  1. 222\sqrt{2} (correct answer)
  2. 424\sqrt{2}
  3. 626\sqrt{2}
  4. 62\sqrt{62}
Explanation: This problem asks us to simplify an expression with radicals by first simplifying each term. We have 72=36⋅2=62\sqrt{72} = \sqrt{36 \cdot 2} = 6\sqrt{2}, 18=9⋅2=32\sqrt{18} = \sqrt{9 \cdot 2} = 3\sqrt{2}, and 8=4⋅2=22\sqrt{8} = \sqrt{4 \cdot 2} = 2\sqrt{2}. Substituting these gives us 62−2(32)+22=62−62+22=226\sqrt{2} - 2(3\sqrt{2}) + 2\sqrt{2} = 6\sqrt{2} - 6\sqrt{2} + 2\sqrt{2} = 2\sqrt{2}. The key error to avoid is trying to combine the numbers under the radicals before simplifying - you must simplify each radical first, then combine like terms. When working with radicals on the SAT, always factor out perfect squares first.

Question 13

Solve the absolute value inequality ∣x+1∣<4|x+1|<4. Express the solution as an interval of real numbers.

  1. (−3,5)(-3,5)
  2. (−5,∞)( -5,\infty)
  3. (−∞,−5)(-\infty,-5)
  4. (−5,3)(-5,3) (correct answer)
Explanation: To solve ∣x+1∣<4|x+1|<4, we use the property that ∣u∣<k|u|<k is equivalent to −k<u<k-k<u<k when k>0k>0. Applying this, we get −4<x+1<4-4 < x+1 < 4. Subtracting 1 from all parts: −4−1<x<4−1-4-1 < x < 4-1, which simplifies to −5<x<3-5 < x < 3. In interval notation, this is (−5,3)(-5,3). The key insight is that an absolute value less than a positive number creates a compound inequality with both upper and lower bounds. A common error is solving only one inequality or using union instead of intersection for the solution.

Question 14

A point PP is on a number line such that its distance from 7 is 3 units. This can be modeled by the equation ∣x−7∣=3|x-7|=3. What are all possible values of xx?

  1. x=4x=4 only
  2. x=−4x=-4 or 1010
  3. x=4x=4 or 1010 (correct answer)
  4. x=10x=10 only
Explanation: This problem asks us to find all points that are 3 units away from 7 on a number line, which translates to solving ∣x−7∣=3|x-7|=3. The absolute value equation splits into two cases: either x−7=3x-7 = 3 or x−7=−3x-7 = -3. From the first case, x=7+3=10x = 7 + 3 = 10. From the second case, x=7−3=4x = 7 - 3 = 4. We can verify: ∣10−7∣=∣3∣=3|10-7| = |3| = 3 ✓ and ∣4−7∣=∣−3∣=3|4-7| = |-3| = 3 ✓. The key insight is that absolute value represents distance, so there are two points equidistant from 7. When solving ∣A∣=k|A| = k where k>0k > 0, always consider both positive and negative cases: A=kA = k and A=−kA = -k.

Question 15

p23=t49\sqrt[3]{p^{2}}=t^{\frac{4}{9}} In the given equation, pp and tt are constants greater than 11. If t=p3nt=p^{3n}, where nn is a constant, what is the value of nn?

  1. 12\dfrac{1}{2} (correct answer)
  2. 32\dfrac{3}{2}
  3. 22
  4. 92\dfrac{9}{2}
Explanation: Rewrite the left side as p2/3p^{2/3}. Substituting t=p3nt=p^{3n} into the right side gives t4/9=(p3n)4/9=p4n/3t^{4/9}=(p^{3n})^{4/9}=p^{4n/3}. Since p>1p>1, the exponents must be equal: 2/3=4n/32/3=4n/3, so 2=4n2=4n and n=1/2n=1/2. A common error is cross-multiplying incorrectly or forgetting to multiply the exponents when raising a power to a power.

Question 16

p34=t18\sqrt[4]{p^{3}}=t^{\frac{1}{8}} In the given equation, pp and tt are constants greater than 11. If t=p2n−4t=p^{2n-4}, where nn is a constant, what is the value of nn?

  1. 33
  2. 55 (correct answer)
  3. 88
  4. 1010
Explanation: The left side is p34p^{\frac{3}{4}}. Substituting t=p2n−4t=p^{2n-4} gives p2n−48p^{\frac{2n-4}{8}}. Setting the exponents equal, 34=2n−48\dfrac{3}{4}=\dfrac{2n-4}{8}, so 2n−4=62n-4=6. Then 2n=102n=10 and n=5n=5. Answering 1010 reports 2n2n rather than nn.

Question 17

p25=t415\sqrt[5]{p^{2}}=t^{\frac{4}{15}} In the given equation, pp and tt are constants greater than 11. If t=p4nt=p^{4n}, where nn is a constant, what is the value of nn?

  1. 38\dfrac{3}{8} (correct answer)
  2. 32\dfrac{3}{2}
  3. 83\dfrac{8}{3}
  4. 66
Explanation: The left side is p25p^{\frac{2}{5}}. Substituting t=p4nt=p^{4n} gives p16n15p^{\frac{16n}{15}}. Setting the exponents equal, 25=16n15\dfrac{2}{5}=\dfrac{16n}{15}. Multiplying both sides by 1515 gives 6=16n6=16n, so n=38n=\dfrac{3}{8}. Here 4n=324n=\dfrac{3}{2}, which is the exponent in t=p4nt=p^{4n}, not the value of nn.

Question 18

p7=t78\sqrt{p^{7}}=t^{\frac{7}{8}} In the given equation, pp and tt are constants greater than 11. If t=p3n−2t=p^{3n-2}, where nn is a constant, what is the value of nn?

  1. 43\dfrac{4}{3}
  2. 2716\dfrac{27}{16}
  3. 116\dfrac{11}{6}
  4. 22 (correct answer)
Explanation: The left side is p72p^{\frac{7}{2}}. Substituting t=p3n−2t=p^{3n-2} gives p7(3n−2)8p^{\frac{7(3n-2)}{8}}. Setting the exponents equal, 72=7(3n−2)8\dfrac{7}{2}=\dfrac{7(3n-2)}{8}. Dividing by 77 and multiplying by 88 gives 3n−2=43n-2=4, so 3n=63n=6 and n=2n=2.

Question 19

If 35x=303\sqrt{5x}=30, what is the value of 4x4x?

  1. 2020
  2. 8080 (correct answer)
  3. 100100
  4. 400400
Explanation: Divide both sides by 33 to get 5x=10\sqrt{5x}=10. Squaring gives 5x=1005x=100, so x=20x=20 and 4x=804x=80. Answering 2020 gives xx itself, and answering 100100 gives 5x5x, the expression under the radical.

Question 20

If 23x=182\sqrt{3x}=18, what is the value of 5x5x?

  1. 135135 (correct answer)
  2. 405405
  3. 540540
  4. 16201620
Explanation: Divide both sides by 22 to get 3x=9\sqrt{3x}=9. Squaring gives 3x=813x=81, so x=27x=27 and 5x=1355x=135. Answering 405405 multiplies 3x=813x=81 by 55 instead of multiplying xx by 55; dropping the 22 altogether gives 3x=3243x=324, x=108x=108, and 5x=5405x=540.