Originally Posted by

**Anemori** i hope im doing it right :P

Solve:

$\displaystyle \frac{1}{(x+1)^2} + \frac{4}{x+1} +4=0 $

if u = x+1 then

$\displaystyle \frac{1}{(u)^2} + \frac {4}{u}+4=0 $

Look for LCD = $\displaystyle u^2 $

$\displaystyle (u)^2 *\frac{1}{(u)^2} + (u)^2 *\frac {4}{u}+ (u)^2*4= 0 $

$\displaystyle \frac{1u^2}{(u)^2} + \frac {4u^2}{u}+ 4u^2 = 0 $

$\displaystyle 1+ 4u+ 4u^2 = 0 $

$\displaystyle 4u^2+ 4u+1 = 0 $

$\displaystyle 4u^2+ 4u = -1 $

I dont really know if im doing this right... please help..