Find the roots of the equation sqrt{3y+1} = s | vedclass

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(D) Given equation: $\sqrt{3y+1} = \sqrt{y-1}$.Squaring both sides,we get:$3y + 1 = y - 1$Rearranging the terms to solve for $y$:$3y - y = -1 - 1$$2y

Vedclass · Accepted Answer

$	ext{None of these}$

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(D) Given equation: $\sqrt{3y+1} = \sqrt{y-1}$.Squaring both sides,we get:$3y + 1 = y - 1$Rearranging the terms to solve for $y$:$3y - y = -1 - 1$$2y = -2$$y = -1$Now,check the validity of the root by substituting $y = -1$ back into the original equation:$LHS$: $\sqrt{3(-1) + 1} = \sqrt{-3 + 1} = \sqrt{-2}$Since $\sqrt{-2}$ is not a real number,$y = -1$ is an extraneous root.Therefore,there is no real root for the given equation.Thus,the correct option is $D$.

Find the roots of the equation $\sqrt{3y+1} = \sqrt{y-1}$.

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