Solve this


$\left(\frac{x}{x+1}\right)^{2}-5\left(\frac{x}{x+1}\right)+6=0,(x \neq-1)$



Given :


Putting $\frac{x}{x+1}=y$, we get:

$y^{2}-5 y+6=0$

$\Rightarrow y^{2}-5 y+6=0$

$\Rightarrow y^{2}-(3+2) y+6=0$

$\Rightarrow y^{2}-3 y-2 y+6=0$

$\Rightarrow y(y-3)-2(y-3)=0$


$\Rightarrow y-3=0$ or $y-2=0$

$\Rightarrow y=3$ or $y=2$

Case I

If $y=3$, we get:


$\Rightarrow x=3(x+1)[$ On cross multiplying $]$

$\Rightarrow x=3 x+3$

$\Rightarrow x=\frac{-3}{2}$

Case II

If $y=2$, we get:


$\Rightarrow x=2(x+1)$

$\Rightarrow x=2 x+2$


$\Rightarrow x=-2$

Hence, the roots of the equation are $\frac{-3}{2}$ and $-2$.


Leave a comment


Click here to get exam-ready with eSaral

For making your preparation journey smoother of JEE, NEET and Class 8 to 10, grab our app now.

Download Now