Advanced Math practice question

Answer: A. x = 5 only

Square both sides: 3x + 1 = x² − 2x + 1, so x² − 5x = 0 and x(x − 5) = 0, giving x = 0 or x = 5. Check each in the ORIGINAL equation: x = 0 gives √1 = 1 but x − 1 = −1, so it fails. x = 5 gives √16 = 4 = 5 − 1. Only x = 5 works.

Why the other answers are wrong

B. x = 0 and x = 5

x = 0 appears after squaring but doesn't satisfy the original equation: a square root can't equal −1.

C. x = 0 only

x = 0 is the extraneous root; the one that survives the check is x = 5.

D. There are no real solutions.

x = 5 is a real solution: √(15 + 1) = 4 and 5 − 1 = 4.

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