Algebra 3 Quiz: Solving Radical Equations
12 questions · exam conditions
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Solving Radical EquationsQuestion 1 of 12

What is the solution set of x+3−2x−1=−1\sqrt{x+3}-\sqrt{2x-1}=-1?

{1}\{1\}
{13}\{13\}
{1,13}\{1,13\}
∅\varnothing
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Algebra 3 Quiz

Algebra 3 Quiz: Solving Radical Equations

Practice Solving Radical Equations in Algebra 3 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 Solving Radical Equations, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra 3.

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

What is the solution set of x+3−2x−1=−1\sqrt{x+3}-\sqrt{2x-1}=-1?

  1. {1}\{1\}
  2. {13}\{13\} (correct answer)
  3. {1,13}\{1,13\}
  4. ∅\varnothing
Explanation: When you see a radical equation, the core danger is extraneous solutions: squaring both sides can create answers that do not actually satisfy the original equation. So your job is not just to solve, but to verify every candidate in the original equation. Start by isolating one radical: x+3=2x−1−1\sqrt{x+3} = \sqrt{2x-1}-1 Wait, better to isolate the more convenient radical. From x+3−2x−1=−1,\sqrt{x+3}-\sqrt{2x-1}=-1, move one radical: x+3+1=2x−1.\sqrt{x+3}+1=\sqrt{2x-1}. Square both sides: x+3+2x+3+1=2x−1,x+3+2\sqrt{x+3}+1=2x-1, so 2x+3=x−5.2\sqrt{x+3}=x-5. Notice this already requires x≥5x\ge 5, because the left side is nonnegative. Square again: 4(x+3)=(x−5)24(x+3)=(x-5)^2 4x+12=x2−10x+254x+12=x^2-10x+25 x2−14x+13=0x^2-14x+13=0 (x−1)(x−13)=0.(x-1)(x-13)=0. So candidates are 11 and 1313. But x=1x=1 fails the condition x≥5x\ge5, and checking the original gives 4−1=1\sqrt4-\sqrt1=1, not −1-1. So 11 is extraneous. Checking x=13x=13: 16−25=4−5=−1.\sqrt{16}-\sqrt{25}=4-5=-1. Thus the solution set is {13}\{13\}. The choice {1}\{1\} comes from forgetting to verify. The choice {1,13}\{1,13\} includes the extraneous root created by squaring. The choice ∅\varnothing incorrectly assumes a negative result is impossible, but one radical can be smaller than the other by 1. Study tip: after solving any radical equation, always substitute your answers back into the original equation. Squaring can introduce false solutions, and checking is the only reliable filter.

Question 2

Solve (3x+4)1/2=x−2(3x+4)^{1/2}=x-2. What is the real solution set?

  1. {0}\{0\}
  2. {0,7}\{0,7\}
  3. {7}\{7\} (correct answer)
  4. ∅\varnothing
Explanation: When you see a radical equation like (3x+4)1/2=x−2(3x+4)^{1/2}=x-2, remember that the radical sign means the principal (nonnegative) square root. Squaring both sides is a valid step, but it can create extraneous solutions, so every candidate must be checked in the original equation. First, square both sides: (3x+4)=(x−2)2(3x+4)=(x-2)^2. Expanding gives 3x+4=x2−4x+43x+4=x^2-4x+4, so x2−7x=0x^2-7x=0, meaning x=0x=0 or x=7x=7. Now check each candidate in the original equation. For x=0x=0, the left side is (3(0)+4)1/2=4=2(3(0)+4)^{1/2}=\sqrt{4}=2, but the right side is 0−2=−20-2=-2. Since 2≠−22 \ne -2, x=0x=0 is extraneous. For x=7x=7, the left side is (21+4)1/2=25=5(21+4)^{1/2}=\sqrt{25}=5, and the right side is 7−2=57-2=5. So the real solution set is {7}\{7\}. The choice {0}\{0\} is wrong because it keeps only the extraneous value of 00, likely from forgetting that 4\sqrt{4} cannot equal −2-2. The choice {0,7}\{0,7\} is wrong because it includes 00 without checking; this is the classic trap of solving by squaring and stopping too soon. The empty set is wrong because it misses the valid solution 77; perhaps the student noticed 00 failed and assumed no solution instead of checking the other candidate. Always isolate the radical, square, solve, and then test every candidate in the original equation. If a candidate fails, discard it — but only after checking all possibilities.

Question 3

A student solves x+3=x+1\sqrt{x+3}=x+1, obtains x=1x=1 and x=−2x=-2 after squaring, and then checks only the squared equation. Which is the best reason that x=−2x=-2 is extraneous?

  1. Substituting x=−2x=-2 makes the radicand x+3x+3 negative, so the square root is undefined.
  2. The value x=−2x=-2 is not a solution of the squared equation; the student made an arithmetic error.
  3. Substituting x=−2x=-2 makes the right side x+1x+1 negative, while x+3\sqrt{x+3} is nonnegative; squaring erased that sign condition. (correct answer)
  4. Any negative candidate obtained by squaring must be rejected, so x=−2x=-2 cannot be valid.
Explanation: Whenever you solve a radical equation by squaring, you are solving a different, wider equation. Squaring can turn false statements into true ones because it destroys sign information: both sides become nonnegative even if they had opposite signs. That is exactly what happened here. Substituting x=−2x=-2 into the original gives −2+3=1=1\sqrt{-2+3}=\sqrt{1}=1 on the left, but x+1=−1x+1=-1 on the right. Since a square root is defined to be nonnegative, 1=−11=-1 is false. However, when you square both sides, you get (−2+3)=(−2+1)2(-2+3)=( -2+1)^2, or 1=11=1, so the candidate survives in the squared equation. Squaring erased the fact that the right side was negative. The choice saying the radicand becomes negative is wrong: x+3=1x+3=1, which is positive, so the square root is defined. The claim that x=−2x=-2 is not a solution of the squared equation is also wrong; it does satisfy the squared equation, so the student did not make an arithmetic error. And the rule "any negative candidate must be rejected" is an overgeneralization — a negative candidate can be valid if both sides of the original equation match; the issue here is sign, not negativity itself. So your strategy: after squaring and solving, always check every candidate in the original equation, not just the squared one. Focus on whether both sides have the same sign. That check is what catches extraneous roots.

Question 4

What is the solution set of x2+3x+5=x+1\sqrt{x^2+3x+5}=x+1?

  1. {−4}\{-4\}
  2. {4}\{4\}
  3. {−4,4}\{-4,4\}
  4. ∅\varnothing (correct answer)
Explanation: Whenever you solve a radical equation, remember that squaring both sides is not reversible: it can create extraneous solutions. Also, the square root symbol means the nonnegative root, so the right-hand side x+1x+1 must be nonnegative too; that is, x≥−1x \ge -1. Square both sides anyway: x2+3x+5=x+1\sqrt{x^2+3x+5}=x+1 x2+3x+5=(x+1)2=x2+2x+1x^2+3x+5=(x+1)^2=x^2+2x+1 [ 3x+5=2x+1 \implies x=-4. But x=−4x=-4 violates the necessary condition x≥−1x\ge -1. Check the original too: the left side becomes 9=3\sqrt{9}=3, while the right side becomes −4+1=−3-4+1=-3. They do not match, so −4-4 is extraneous. Therefore no real number satisfies the equation; the solution set is empty. The choice {−4}\{-4\} is the classic trap: it comes from squaring and solving but forgetting to check whether the candidate actually works. The choice {−4,4}\{-4,4\} is also impossible because squaring produced exactly one candidate, so 44 cannot be added as a solution. And the choice {4}\{4\} fails direct substitution:when x=4x=4, left side is 33\sqrt{33}, not 55. These errors usually come from sign mistakes or from treating extraneous roots as valid. Your takeaway: in any radical equation, after squaring, always test every candidate in the original equation. Even better, before squaring, impose the condition that the non-radical side must be nonnegative;this catches most extraneous solutions immediately.

Question 5

What is the real solution of x+5+x−2=4\sqrt{x+5}+\sqrt{x-2}=4?

  1. {20964}\left\{\frac{209}{64}\right\} (correct answer)
  2. {8164}\left\{\frac{81}{64}\right\}
  3. {98}\left\{\frac{9}{8}\right\}
  4. ∅\varnothing
Explanation: When you see a radical equation like x+5+x−2=4\sqrt{x+5}+\sqrt{x-2}=4, your first move should be to isolate a radical, then square both sides. But because squaring can introduce extraneous solutions, always check your final answer in the original equation. Start by isolating: x+5=4−x−2\sqrt{x+5}=4-\sqrt{x-2}. Squaring gives x+5=16−8x−2+x−2x+5=16-8\sqrt{x-2}+x-2. Simplify: x+5=x+14−8x−2x+5=x+14-8\sqrt{x-2}, so −9=−8x−2-9=-8\sqrt{x-2}, meaning x−2=98\sqrt{x-2}=\frac{9}{8}. Squaring again yields x−2=8164x-2=\frac{81}{64}, so x=20964x=\frac{209}{64}. Checking in the original equation confirms it works: 20964+5+20964−2=238+98=4\sqrt{\frac{209}{64}+5}+\sqrt{\frac{209}{64}-2}=\frac{23}{8}+\frac{9}{8}=4. The choice 8164\frac{81}{64} is a trap: it is the value of x−2x-2, not the value of xx. You must add the 2 back to solve for xx. Similarly, 98\frac{9}{8} is the value of x−2\sqrt{x-2} after the first squaring step, not the final solution for xx. Finally, ∅\varnothing is incorrect because the solution 20964\frac{209}{64} is real and satisfies the original equation; there is no extraneous root here. Your study tip: whenever you square a radical equation, finish by substituting your candidates back into the original equation. That check is what separates valid solutions from extraneous ones.

Question 6

Solve 2x+3=x−4\sqrt{2x+3}=x-4. Which of the following is the correct real solution set?

  1. {5−23}\{5-2\sqrt{3}\}
  2. {5+23}\{5+2\sqrt{3}\} (correct answer)
  3. {5−23, 5+23}\{5-2\sqrt{3},\,5+2\sqrt{3}\}
  4. ∅\varnothing
Explanation: When you see a radical equation like 2x+3=x−4\sqrt{2x+3}=x-4, your first instinct may be to square both sides. That is correct, but squaring can create extraneous solutions—numbers that satisfy the squared equation but not the original. You must always check your answers against the original equation. Squaring gives 2x+3=(x−4)22x+3=(x-4)^2, which simplifies to x2−10x+13=0x^2-10x+13=0. Using the quadratic formula, x=5±23x=5\pm 2\sqrt{3}. Now test each candidate. The value 5−23≈1.545-2\sqrt{3}\approx 1.54 makes the right side x−4x-4 negative. But the left side, 2x+3\sqrt{2x+3}, is always nonnegative, so it cannot equal a negative number. Thus 5−235-2\sqrt{3} is extraneous. The value 5+23≈8.465+2\sqrt{3}\approx 8.46 makes both sides nonnegative, and substituting it back satisfies the original equation. So the correct real solution set is {5+23}\{5+2\sqrt{3}\}. The choice {5−23}\{5-2\sqrt{3}\} represents the trap of stopping after solving the squared equation without checking. The choice containing both values includes the extraneous root, which many students select when they forget the domain condition that x−4≥0x-4\ge 0. The empty set is incorrect because the original equation does have one valid real solution. A reliable strategy: whenever you square both sides, immediately remember that every candidate must be checked. Also, before solving, note that the right side must be nonnegative, so x≥4x\ge 4. That single condition quickly eliminates the false root.

Question 7

What is the solution set of x+4=x−2\sqrt{x+4}=\sqrt{x}-2?

  1. {0}\{0\}
  2. {4}\{4\}
  3. {0,4}\{0,4\}
  4. ∅\varnothing (correct answer)
Explanation: Whenever you solve an equation involving square roots, the key is to remember that squaring both sides can create extraneous solutions. So after solving, you must check every candidate in the original equation. Here, set y=xy=\sqrt{x}, so y≥0y\ge 0. The equation becomes y2+4=y−2.\sqrt{y^2+4}=y-2. The right side y−2y-2 must be nonnegative, so y≥2y\ge 2. Squaring both sides gives y2+4=y2−4y+4⇒0=−4y⇒y=0.y^2+4=y^2-4y+4 \quad\Rightarrow\quad 0=-4y \quad\Rightarrow\quad y=0. But y=0y=0 contradicts y≥2y\ge 2, so there is no solution. Checking the proposed answers confirms this: {0}\{0\} comes from the extraneous result y=0y=0, but substituting x=0x=0 gives 4=2\sqrt{4}=2 on the left and 0−2=−2\sqrt{0}-2=-2 on the right. {4}\{4\} fails too, since 8≠0\sqrt{8}\neq 0. And {0,4}\{0,4\} just combines two non-solutions. Therefore the solution set is ∅\varnothing. Study tip: after squaring a radical equation, always plug your answers back into the original equation, or check the required sign condition before squaring. That will quickly eliminate extraneous roots like x=0x=0 here.

Question 8

Solve 2x+1=−x\sqrt{2x+1}=-x. What is the real solution set?

  1. {1+2}\{1+\sqrt{2}\}
  2. {1−2}\{1-\sqrt{2}\} (correct answer)
  3. {1+2, 1−2}\{1+\sqrt{2},\,1-\sqrt{2}\}
  4. ∅\varnothing
Explanation: Whenever you solve a radical equation like 2x+1=−x\sqrt{2x+1}=-x, remember that the square root symbol means the nonnegative root. So the right-hand side, −x-x, must also be nonnegative, which means x≤0x \le 0. Also, the radicand requires 2x+1≥02x+1 \ge 0, so x≥−12x \ge -\frac12. Any valid solution must lie in [−12,0][-\tfrac12, 0]. Squaring both sides gives 2x+1=x22x+1 = x^2, or x2−2x−1=0x^2 - 2x - 1 = 0. By the quadratic formula, x=1±2x = 1 \pm \sqrt{2}. Now test these against the original equation. The value 1+2≈2.4141+\sqrt{2} \approx 2.414 is positive, so it violates the requirement that −x-x be nonnegative; it is an extraneous root introduced by squaring. The value 1−2≈−0.4141-\sqrt{2} \approx -0.414 is in the valid interval, and checking: 2(1−2)+1=−(1−2)\sqrt{2(1-\sqrt2)+1} = -(1-\sqrt2) works. So the real solution set is {1−2}\{1-\sqrt{2}\}. The choice listing 1+21+\sqrt{2} alone misses the domain check. The choice with both values includes the extraneous positive root. The empty set is the trap if you assume −x-x is always negative; actually, when xx is negative, −x-x is positive, so a radical can equal it. Your study tip: after squaring a radical equation, always check every candidate in the original equation, and use sign restrictions before solving to filter out extraneous answers quickly.

Question 9

Solve x+2+2=3x+4\sqrt{x+2}+2=\sqrt{3x+4}. What is the real solution set?

  1. {−1}\{-1\}
  2. {−1,7}\{-1,7\}
  3. {7}\{7\} (correct answer)
  4. ∅\varnothing
Explanation: When you see a radical equation, remember that squaring both sides can create extraneous solutions, so every candidate you find must be checked in the original equation. First, note the domain requires x≥−43x \ge -\frac43. Start with x+2+2=3x+4.\sqrt{x+2}+2=\sqrt{3x+4}. Square both sides: x+2+4x+2+4=3x+4x+2+4\sqrt{x+2}+4=3x+4 which simplifies to 4x+2=2x−2.4\sqrt{x+2}=2x-2. Divide by 2: 2x+2=x−1.2\sqrt{x+2}=x-1. Squaring again gives 4(x+2)=(x−1)2,4(x+2)=(x-1)^2, so x2−6x−7=0,x^2-6x-7=0, and therefore x=7x=7 or x=−1x=-1. Now check both. For x=7x=7, 9+2=3+2=5=25,\sqrt{9}+2=3+2=5=\sqrt{25}, so 77 works. For x=−1x=-1, 1+2=3≠1=1,\sqrt{1}+2=3 \neq \sqrt{1}=1, so −1-1 is extraneous. Thus the real solution set is {7}\{7\}. The choice {−1}\{-1\} traps students who keep the extraneous root after solving the squared equation. The choice {−1,7}\{-1,7\} lists both algebraic solutions but fails to verify them in the original equation. The empty set would be chosen by someone who rejects −1-1 but overlooks that 77 checks correctly. Always substitute every candidate back into the original radical equation; if an intermediate step requires a nonnegative side, like 2x+2=x−12\sqrt{x+2}=x-1, that condition can also help you spot extraneous roots quickly.

Question 10

Solve x2+4x+4=x+6\sqrt{x^2+4x+4}=x+6. What is the real solution set?

  1. {−4}\{-4\} (correct answer)
  2. {−2}\{-2\}
  3. {−4,−2}\{-4,-2\}
  4. ∅\varnothing
Explanation: When you see a square root containing a perfect-square quadratic, your first thought should be: a2=∣a∣\sqrt{a^2}=|a|. Here, x2+4x+4=(x+2)2x^2+4x+4=(x+2)^2, so the equation becomes (x+2)2=x+6⇒∣x+2∣=x+6.\sqrt{(x+2)^2}=x+6 \quad \Rightarrow \quad |x+2|=x+6. Now split into cases. If x+2≥0x+2\ge 0, then x+2=x+6x+2=x+6, which gives 2=62=6, impossible. If x+2<0x+2<0, then −(x+2)=x+6-(x+2)=x+6, so −x−2=x+6⇒−2x=8⇒x=−4.-x-2=x+6 \quad \Rightarrow \quad -2x=8 \quad \Rightarrow \quad x=-4. Check x=−4x=-4: the left side is 16−16+4=2\sqrt{16-16+4}=2, and the right side is −4+6=2-4+6=2. So {−4}\{-4\} is the real solution set. The choice {−2}\{-2\} fails because plugging in −2-2 gives 00 on the left but 44 on the right; it only makes the radicand zero, not the whole equation true. The choice {−4,−2}\{-4,-2\} includes the valid −4-4 but repeats that same extraneous −2-2. The choice ∅\varnothing comes from mistakenly simplifying (x+2)2\sqrt{(x+2)^2} to x+2x+2 instead of ∣x+2∣|x+2|; that leads to x+2=x+6x+2=x+6, which has no solution, but the absolute-value equation does have one. Always rewrite perfect square\sqrt{\text{perfect square}} as an absolute value, and check every candidate in the original equation before finalizing your solution set.

Question 11

Which solution set is correct for 2x+13=x+1\sqrt[3]{2x+1}=x+1?

  1. {0}\{0\}
  2. {−3+52}\left\{\frac{-3+\sqrt{5}}{2}\right\}
  3. {−3±52}\left\{\frac{-3\pm\sqrt{5}}{2}\right\}
  4. {0,−3±52}\left\{0,\frac{-3\pm\sqrt{5}}{2}\right\} (correct answer)
Explanation: Whenever you see a cube-root equation, your first move should be to cube both sides. Unlike squaring, cubing is a one-to-one operation, so it does not introduce extraneous solutions. But after you cube, you still need to solve the resulting polynomial completely. For 2x+13=x+1\sqrt[3]{2x+1}=x+1, cube both sides to get 2x+1=(x+1)3=x3+3x2+3x+1.2x+1=(x+1)^3=x^3+3x^2+3x+1. Subtract 2x+12x+1 from both sides: 0=x3+3x2+x=x(x2+3x+1).0=x^3+3x^2+x=x(x^2+3x+1). So the solutions come from two factors: x=0x=0, or x2+3x+1=0x^2+3x+1=0. The quadratic gives x=−3±52.x=\frac{-3\pm\sqrt{5}}{2}. Plugging each into the original confirms all three work. Now look at the wrong answers. The set {0}\{0\} is incomplete: it comes from setting only the factor xx equal to zero and ignoring the quadratic factor. The set containing only −3+52\frac{-3+\sqrt{5}}{2} misses both 00 and the other quadratic root; it might come from mistakenly taking only one sign of the square root. The set {−3±52}\left\{\frac{-3\pm\sqrt{5}}{2}\right\} correctly lists both quadratic roots but still misses the solution x=0x=0, probably because the factor xx was divided away or overlooked entirely. Study tip: after cubing, always factor the polynomial completely. If you divide out a factor—especially xx—you lose solutions. Set everything equal to zero, factor, and solve each factor separately.

Question 12

A student correctly squares x+6=x+4\sqrt{x+6}=x+4 and obtains the candidates x=−5x=-5 and x=−2x=-2. Which value or values are extraneous?

  1. Only x=−5x=-5 (correct answer)
  2. Only x=−2x=-2
  3. Both x=−5x=-5 and x=−2x=-2
  4. Neither x=−5x=-5 nor x=−2x=-2
Explanation: Whenever you square both sides of an equation, you can create extraneous solutions: squaring is not reversible, because a=ba=b implies a2=b2a^2=b^2, but a2=b2a^2=b^2 does not imply a=ba=b. So after solving the squared equation, you must plug each candidate back into the original radical equation. For x=−5x=-5: −5+6=1=1,\sqrt{-5+6}=\sqrt{1}=1, but the right side is −5+4=−1.-5+4=-1. Since 1≠−11 \ne -1, x=−5x=-5 fails and is extraneous. For x=−2x=-2: −2+6=4=2,\sqrt{-2+6}=\sqrt{4}=2, and the right side is −2+4=2.-2+4=2. This matches, so x=−2x=-2 is valid. Thus "Only x=−5x=-5" is correct. "Only x=−2x=-2" mistakes the valid solution for the extraneous one. "Both x=−5x=-5 and x=−2x=-2" assumes squaring always invalidates both candidates, but only the one that fails the original check is extraneous. "Neither x=−5x=-5 nor x=−2x=-2" misses that squaring can introduce an extra solution, and x=−5x=-5 is exactly such a false solution because it makes the original equation have 11 on the left and −1-1 on the right. On any radical equation, make it a habit to substitute your candidates into the original equation. If a candidate satisfies the original equation, keep it; if not, discard it as extraneous.