A student solves √x = x − 2 by squaring both sides and obtains x = 1 and x = 4. After substituting back into the original equation, which conclusion is correct?
ABoth x = 1 and x = 4 are valid solutions
BOnly x = 4 is valid; x = 1 is extraneous because √1 = 1 but 1 − 2 = −1, so they are not equal
COnly x = 1 is valid; x = 4 is extraneous because it is farther from zero
DBoth are extraneous; squaring both sides always invalidates all solutions
Substituting x = 1: √1 = 1, but 1 − 2 = −1. Since 1 ≠ −1, x = 1 fails the check and is extraneous. Substituting x = 4: √4 = 2, and 4 − 2 = 2. ✓ Squaring introduced x = 1 because it eliminated the sign constraint — the step (√x)² = (x−2)² is satisfied when √x = x−2 OR when √x = −(x−2). Checking restores the original constraint.
Question 2 Multiple Choice
Why do cube root equations never produce extraneous solutions when you cube both sides, whereas square root equations routinely produce them when you square both sides?
ABecause cube roots can be negative, which makes the algebra more forgiving
BBecause cubing is a one-to-one function — each output corresponds to exactly one input — so no false equations are created by the operation
CBecause cube root equations are always simpler and don't require checking
DBecause squaring is allowed on equations but cubing is not, so different rules apply
Squaring loses sign information: both 3² and (−3)² equal 9, so squaring both sides of an equation can create a true equation from a false one (e.g., 3 ≠ −3, but 9 = 9). Cubing is one-to-one: only 3³ = 27, and only (−3)³ = −27. So cubing cannot turn a false equation into a true one, and no extraneous solutions are introduced. This is the deeper reason the checking step is essential for even-index roots but not for odd-index roots.
Question 3 True / False
Squaring both sides of a radical equation is a fully reversible algebraic step that preserves most solutions without introducing new ones.
TTrue
FFalse
Answer: False
Squaring is not reversible in general because it discards sign information. Both a = b and a = −b lead to a² = b² after squaring, so squaring can produce a true equation from a false one. This is exactly how extraneous solutions arise: squaring turns the unsatisfiable statement '√x = −3' into the satisfiable 'x = 9', which passes algebra but fails the original equation. The solving process is only complete after checking all solutions in the original equation.
Question 4 True / False
Checking solutions in the original radical equation after solving is mathematically necessary, not merely a good habit, because the squaring step can produce solutions to a related but different equation.
TTrue
FFalse
Answer: True
When you square both sides, you solve a new equation — one that is satisfied both when the two sides are equal and when they are negatives of each other. Some solutions to this new equation will not satisfy the original. The check is the mechanism that filters these out. Skipping it means you may report a number that does not actually satisfy the original equation, which is a mathematical error, not just sloppiness.
Question 5 Short Answer
Explain why the equation √(x + 4) = −2 has no solution, and what happens if you square both sides without thinking.
Think about your answer, then reveal below.
Model answer: The square root function always returns a non-negative value, so √(x + 4) ≥ 0 for all valid x. It can never equal −2. If you square both sides without recognizing this, you get x + 4 = 4, so x = 0. But checking: √(0 + 4) = √4 = 2 ≠ −2. The solution x = 0 is extraneous — introduced because squaring eliminated the sign, making the impossible statement √(x+4) = −2 look solvable. This example shows why checking is mandatory: the algebra produces a number, but that number is not a solution.
The key insight is that √x ≥ 0 always. Any equation of the form √(...) = [negative number] has no solution, and the checking step will always catch this. Squaring both sides of such an equation will produce an extraneous solution every time, because squaring −2 and squaring +2 give the same result.