Solve {l}{text{Findtherootsoftheequation,}sqrt{3y+1}=sqrt{y-1}}{text{Wehave,onsquaring.}}{text{3y}+1=y-1Rightarrow2y=-2Rightarrowy=-1}

$$\left. \begin{array} { l } { \text { Find the roots of the equation, } \sqrt { 3 y + 1 } = \sqrt { y - 1 } } \\ { \text { We have, on squaring. } } \\ { \text { 3y } + 1 = y - 1 \Rightarrow 2 y = - 2 \Rightarrow y = - 1 } \end{array} \right.$$

Solve for y

$$y=-1$$

Solution Steps

Square both sides of the equation.

$$\left(\sqrt{3y+1}\right)^{2}=\left(\sqrt{y-1}\right)^{2}$$

Calculate \sqrt{3y+1} to the power of 2 and get 3y+1.

$$3y+1=\left(\sqrt{y-1}\right)^{2}$$

Calculate \sqrt{y-1} to the power of 2 and get y-1.

$$3y+1=y-1$$

Subtract y from both sides.

$$3y+1-y=-1$$

Combine 3y and -y to get 2y.

$$2y+1=-1$$

Subtract 1 from both sides.

$$2y=-1-1$$

Subtract 1 from -1 to get -2.

$$2y=-2$$

Divide both sides by 2.

$$y=\frac{-2}{2}$$

Divide -2 by 2 to get -1.

$$y=-1$$

Substitute -1 for y in the equation \sqrt{3y+1}=\sqrt{y-1}.

$$\sqrt{3\left(-1\right)+1}=\sqrt{-1-1}$$

Simplify. The value y=-1 satisfies the equation.

$$i\times 2^{\frac{1}{2}}=i\times 2^{\frac{1}{2}}$$

Equation \sqrt{3y+1}=\sqrt{y-1} has a unique solution.

$$y=-1$$