# How do you solve 4/(b+1)= 1 + 2/(2b-1)?

Oct 11, 2015

The given equation has no Real solutions
(see below for Complex solution values)

#### Explanation:

Given
$\textcolor{w h i t e}{\text{XXX}} \frac{4}{b + 1} = 1 + \frac{2}{2 b - 1}$

Multiply both sides by $\left(b + 1\right) \left(2 b - 1\right)$
$\textcolor{w h i t e}{\text{XXX}} 4 \left(2 b - 1\right) = \left(b + 1\right) \left(2 b - 1 + 2\right)$

Simplify:
$\textcolor{w h i t e}{\text{XXX}} 8 b - 4 = \left(b + 1\right) \left(2 b + 1\right)$

$\textcolor{w h i t e}{\text{XXX}} 8 b - 4 = 2 {b}^{2} + 3 b + 1$

Re-write in standard parabolic form
$\textcolor{w h i t e}{\text{XXX}} 2 {b}^{2} - 5 b + 5 = 0$

$\textcolor{w h i t e}{\text{XXX}} b = \frac{5 \pm \sqrt{{5}^{2} - 4 \left(2\right) \left(5\right)}}{2 \left(2\right)}$

$\textcolor{w h i t e}{\text{XXX}} b = \frac{5 \pm \sqrt{- 15}}{4}$
with no Real roots (because of the negative square root value)
but
Complex roots at $\frac{5}{4} \pm \frac{\sqrt{15}}{4} i$

If we re-write
$\textcolor{w h i t e}{\text{XXX}} \frac{4}{b + 1} = \frac{2}{2 b - 1} + 1$
as
$\textcolor{w h i t e}{\text{XXX}} 4 \left(b + 1\right) - \frac{2}{2 b - 1} - 1 = 0$
then examine the graph of left side:
graph{4/(x+1)-2/(2x-1)-1 [-10, 10, -5, 5]}

We see that at no point does it touch the X-axis (i.e. it is never equal to $0$)