1. ## u substitution

i hope im doing it right :P

Solve:

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

if u = x+1 then

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

Look for LCD = $u^2$

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

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

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

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

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

2. You don't multiply 1/u^2 by u^2, you multiply the other two terms by u^2.

3. like this?

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

4. Originally Posted by Anemori
i hope im doing it right :P

Solve:

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

if u = x+1 then

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

Look for LCD = $u^2$

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

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

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

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

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

$\frac{1}{u^2} + \frac{4}{u} + 4 = 0$.

Multiply both sides of the equation by $u^2$.

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

$1 + 4u + 4u^2 = 0$.

Now you have a Quadratic equation - use the Quadratic Formula or some other method to solve for $u$.

i hope this is right way...

$x = \frac {-4 +- \sqrt4^2- 4(4)(1)}{2(4)}$

$x = \frac {-4 +- \sqrt16- (16)}{8}$

$x = \frac {-4 +- \sqrt0}{8}$

$x = \frac {-4 +- 0}{8}$

$x = \frac {-4 +- 0}{8}$

$x = -\frac{1}{2}$

Is that right?

6. Originally Posted by Anemori

i hope this is right way...

$x = \frac {-4 +- \sqrt4^2- 4(4)(1)}{2(4)}$

$x = \frac {-4 +- \sqrt16- (16)}{8}$

$x = \frac {-4 +- \sqrt0}{8}$

$x = \frac {-4 +- 0}{8}$

$x = \frac {-4 +- 0}{8}$

$x = -\frac{1}{2}$

Is that right?
If you have done what Prove it told you to do, then it is correct.

7. How do you check if it is right?

8. Originally Posted by Anemori
How do you check if it is right?
Plug it back in.

9. $1+ 4u+ 4u^2$ factors rather easily as $(1+ 2u)^2$. That gives $u= -\frac{1}{2}$, not x! Since you took u= x+ 1 you now know that $x+1= -\frac{1}{2}$.