How do you solve 1+ 6( 2x + 3) ^- 1+ 8( 2x + 3) ^- 2= 0?

Sep 13, 2016

The Soln. $: x = - \frac{7}{2} , \mathmr{and} , x = - \frac{5}{2}$

Explanation:

$1 + 6 {\left(2 x + 3\right)}^{- 1} + 8 {\left(2 x + 3\right)}^{- 2} = 0$

Multiplying by, ${\left(2 x + 3\right)}^{2}$, we get,

${\left(2 x + 3\right)}^{2} + 6 \left(2 x + 3\right) + 8 = 0$. Letting, $\left(2 x + 3\right) = y$, we get,

${y}^{2} + 6 y + 8 = 0$

$\therefore \left(y + 4\right) \left(y + 2\right) = 0$

$\therefore y = - 4 , \mathmr{and} , y = - 2$

$\therefore 2 x + 3 = - 4 , \mathmr{and} , 2 x + 3 = - 2$

$\therefore 2 x = - 7 , \mathmr{and} , 2 x = - 5$

$\therefore x = - \frac{7}{2} , \mathmr{and} , x = - \frac{5}{2}$

These roots satisfy the given eqn.

Hence, the Soln. $: x = - \frac{7}{2} , \mathmr{and} , x = - \frac{5}{2}$

Sep 13, 2016

$x = - \frac{7}{2} = - 3 \frac{1}{2} \text{ or } x = - \frac{5}{2} = - 2 \frac{1}{2}$

Explanation:

Recall: ${x}^{-} m = \frac{1}{x} ^ m$

Get rid of the negative indices first

$1 + 6 {\left(2 x + 3\right)}^{-} 1 + 8 {\left(2 x + 3\right)}^{-} 2 = 0$

$1 + \frac{6}{2 x + 3} + \frac{8}{2 x + 3} ^ 2 = 0$

As it is an equation, we can get rid of the denominators by multiplying through by the LCD which in this case is $\textcolor{red}{{\left(2 x + 3\right)}^{2}}$

$1 \times \textcolor{red}{{\left(2 x + 3\right)}^{2}} + \frac{6 \times \textcolor{red}{{\left(2 x + 3\right)}^{2}}}{2 x + 3} + \frac{8 \times \textcolor{red}{{\left(2 x + 3\right)}^{2}}}{2 x + 3} ^ 2 = 0$

$\textcolor{red}{{\left(2 x + 3\right)}^{2}} + 6 \textcolor{red}{\left(2 x + 3\right)} + 8 = 0$

$4 {x}^{2} + 12 x + 9 + 12 x + 18 + 8 = 0$

$4 {x}^{2} + 24 x + 35 = 0 \text{ } \leftarrow$ find factors

$\left(2 x + 7\right) \left(2 x + 5\right) = 0$

If $2 x + 7 = 0 \rightarrow x = - \frac{7}{2}$

If $2 x + 5 = 0 \rightarrow x = - \frac{5}{2}$