Question

`(x/(x+1))^2-5(x/(x+1)+6=0,x≠b,a` 

Answer 1

`(x/(x+1))^2-5(x/(x+1))+6=0` 

Putting `x/(x+1)=y`, We get: 

⇒`y^2-5y+6=0` 

⇒`y^2-5y+6=0` 

⇒`y^2-(3+2)y+6=0`  

⇒`y^2-3y-2y+6=0` 

⇒`y(y-3)-2(y-3)=0` 

⇒`(y-3)(y-2)=0` 

⇒`y-3=0  or  y-2=0` 

`y=3  or  y=2`  Case I:
If y = 3, we get 

⇒`x/(x+1)=3`  

⇒x=3(x+1)           [On cross multiplying 

⇒`x=3x+3` 

⇒`x=(-3)/2` 

Case II:
If y =2, we get: 

⇒`x/(x+1)=2` 

⇒`x=2(x+1) `

⇒`x=2x+2` 

⇒`-x=2` 

⇒`x=-2` 

Hence, the roots of the equation are `-3/2` and -2