# How do you solve 2/(x^2+2x+1)-3/(x+1)=4?

Feb 4, 2015

You can write the first denominator as:
${x}^{2} + 2 x + 1 = \left(x + 1\right) \left(x + 1\right) = {\left(x + 1\right)}^{2}$
So yo get:
$\frac{2}{{\left(x + 1\right)}^{2}} - \frac{3}{x + 1} = 4$

You find a common denominator for both sides and "rearrange" the numerators to adapt to this, so you get:

$\frac{2}{{\left(x + 1\right)}^{2}} - \frac{3 \cdot \left(x + 1\right)}{{\left(x + 1\right)}^{2}} = \frac{4 \cdot {\left(x + 1\right)}^{2}}{x + 1} ^ 2$

Multiplying both sides by ${\left(x + 1\right)}^{2}$ you get:

$2 - 3 \left(x + 1\right) = 4 {\left(x + 1\right)}^{2}$
$2 - 3 \left(x + 1\right) = 4 \left({x}^{2} + 2 x + 1\right)$
$2 - 3 x - 3 = 4 {x}^{2} + 8 x + 4$
$4 {x}^{2} - 11 x + 5 = 0$
Which is a second degree equation.

Hope it helps