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

**Neverquit** · April 9th 2010, 09:54 PM

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?

**harish21** · April 9th 2010, 10:03 PM

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? ?

**Neverquit** · April 9th 2010, 10:06 PM

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.

**chiph588@** · April 9th 2010, 10:06 PM

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.

**harish21** · April 9th 2010, 10:09 PM

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!

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