# Thread: [SOLVED] Can't solve this with quotient rule!!

1. ## [SOLVED] Can't solve this with quotient rule!!

I have this function Y=(X^2+2)/(x-1)
and get this Y=2x(x-1)-(x^2+2)/( x-1)^2.

When I simplify to quadratic form I don’t get….x=1-sqrt3,1+sqrt3
Can anyone help?

2. Originally Posted by Neverquit
I have this function Y=(X^2+2)/(x-1)
and get this Y=2x(x-1)-(x^2+2)/( x-1)^2.

When I simplify to quadratic form I don’t get….x=1-sqrt3,1+sqrt3
Can anyone help?

your differentiation looks good. What are you trying to do after that? ?

3. Originally Posted by harish21
Did you mean solving for $Y' = 0$?
Yes I tried to solve for x when f=0. But I don't get x=1-sqrt3,1+sqrt3.

4. Originally Posted by Neverquit
I have this function Y=(X^2+2)/(x-1)
and get this Y=2x(x-1)-(x^2+2)/( x-1)^2.

When I simplify to quadratic form I don’t get….x=1-sqrt3,1+sqrt3
Can anyone help?
I assume you're wanting to solve $y'=0$

$y' = \frac{2x(x-1)-(x^2+2)}{(x-1)^2}=0\implies 2x(x-1)-(x^2+2)=0\implies x^2-2x-2=0$

Now just apply the quadratic formula and we're done.

5. As chip588 has stated above, you end up with

$x^2-2x-2=0$

This is a quadratic equation of the form $ax^2+bx+c=0$

where, $a = 1$

$b = -2$

$c= -2$

solve for x with this formula

$x = \frac {-b \pm \sqrt{b^{2}-4ac}}{2a}$

With this formula, you will get the answer you are looking for. Give it a shot!