The roots of the equation $\sqrt{3y + 1} = \sqrt{y - 1}$ are?
- $0, - 1$
- $2, 3$
- $2, 1$
-
None of these
Reveal answer
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Correct answer
Explanation
Squaring both sides gives 3y + 1 = y - 1, which leads to 2y = -2, so y = -1. Checking y = -1 in the original equation: sqrt(3(-1)+1) = sqrt(-2), which is undefined in real numbers. Thus, there are no real solutions.
AI explanation
Squaring both sides of the equation gives 3y + 1 = y - 1, which simplifies to 2y = -2 and yields the root y = -1. Substituting y = -1 back into the original equation results in the square root of a negative number, making it an extraneous root. Since the only derived root is rejected, the original equation has no valid roots.