$2x-2=\sqrt{3x^2+13}$ . Which value(s) of $x$ satisfies the equation above? I. $-1$ II. $4$ III. $9$

Answer Choices

Correct answer marked below

III

Correct

I & II

I & III

I, II, & III

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Comments

BoniJun 18

Classic trap question. I nearly forgot to test the candidates in the original equation instead of just squaring and trusting the results.

TanishaJul 1

This is why I always plug them back into the original. Extraneous roots love square-root equations.

HardikApr 19

I don't understand why III is in the answer choices at all. I squared everything correctly but only got two values. Can someone explain how 9 could even seem possible?

NishthaApr 27

Probably a distractor for people who mis-expand the trinomial. I almost picked it until I plugged x = 9 into the left side.

AudreyAug 22

Wait, this felt harder than the difficulty suggests. I got tangled in the squaring and almost forgot to check the domain condition.

FizaSep 2

The trap is real. I got the right answer but only after wasting 3 minutes. On test day I'd flag and move on.

DanielAug 12

This is a solid example of why you square both sides, solve, then check for extraneous solutions. Thanks for the practice.

What this question tests

What it tests

Your handling of square and higher roots, including simplifying radicals.

Common trap

Forgetting the ± when taking a square root to solve an equation, or assuming √(a²) = a.

2x−2=3x2+132x-2=\sqrt{3x^2+13}2x−2=3x2+13​. Which value(s) of xxx satisfies the equation above? I. −1-1−1 II. 444 III. 999