Advanced Math practice question

Answer: B. {4}

Isolate the radical: √(x + 5) = x − 1. Squaring gives x + 5 = x² − 2x + 1, so x² − 3x − 4 = 0, or (x − 4)(x + 1) = 0, giving candidates x = 4 and x = −1. Check each in the ORIGINAL equation. For x = 4: 4 − √9 = 4 − 3 = 1 ✓. For x = −1: −1 − √4 = −1 − 2 = −3 ≠ 1 ✗ (the principal square root is +2, not −2). So the only solution is 4. Choice A is the extraneous root, selected by a student who verifies in √(x + 5) = x − 1 while allowing the radical to equal a negative value (−2 = −2). Choice C is the result of solving the squared quadratic correctly but never checking for extraneous solutions. Choice D reflects the misconception that squaring both sides automatically invalidates every candidate, or an arithmetic slip such as x + 5 = x² − 1 producing non-real roots.

Why the other answers are wrong

A. {−1}

x = −1 is the extraneous root: it only works if you let √4 = −2, but the principal square root is +2, so −1 − 2 = −3, not 1.

C. {−1, 4}

You solved the squared quadratic correctly but skipped the check in the original equation; squaring can create extraneous roots, and x = −1 fails.

D. There are no real solutions.

Squaring both sides doesn't invalidate every candidate — check them: x = 4 gives 4 − √9 = 1, a genuine solution.

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