# How do you solve [(x^2+5x) / (x +2)] + 10x = -24?

Nov 5, 2017

See a solution process below:

#### Explanation:

First, subtract $\textcolor{red}{10 x}$ from each side of the equation to isolate the fraction while keeping the equation balanced:

$\left[\frac{{x}^{2} + 5 x}{x + 2}\right] + 10 x - \textcolor{red}{10 x} = - 24 - \textcolor{red}{10 x}$

$\left[\frac{{x}^{2} + 5 x}{x + 2}\right] + 0 = - 24 - 10 x$

$\left[\frac{{x}^{2} + 5 x}{x + 2}\right] = - 10 x - 24$

Next, multiply each side of the equation by $\textcolor{red}{\left(x + 2\right)}$ to eliminate the fraction while keeping the equation balanced:

$\textcolor{red}{\left(x + 2\right)} \left[\frac{{x}^{2} + 5 x}{x + 2}\right] = \textcolor{red}{\left(x + 2\right)} \left(- 10 x - 24\right)$

$\cancel{\textcolor{red}{\left(x + 2\right)}} \left[\frac{{x}^{2} + 5 x}{\textcolor{red}{\cancel{\textcolor{b l a c k}{x + 2}}}}\right] = \left(\textcolor{red}{x} \cdot - 10 x\right) + \left(\textcolor{red}{x} \cdot - 24\right) + \left(\textcolor{red}{2} \cdot - 10 x\right) + \left(\textcolor{red}{2} \cdot - 24\right)$

${x}^{2} + 5 x = - 10 {x}^{2} - 24 x - 20 x - 48$

${x}^{2} + 5 x = - 10 {x}^{2} - 44 x - 48$

Then, add $\textcolor{red}{10 {x}^{2}}$ and $\textcolor{b l u e}{44 x}$ and $\textcolor{\mathmr{and} a n \ge}{48}$ to each side of the equation to put the equation in standard form:

${x}^{2} + \textcolor{red}{10 {x}^{2}} + 5 x + \textcolor{b l u e}{44 x} + \textcolor{\mathmr{and} a n \ge}{48} = - 10 {x}^{2} + \textcolor{red}{10 {x}^{2}} - 44 x + \textcolor{b l u e}{44 x} - 48 + \textcolor{\mathmr{and} a n \ge}{48}$

$1 {x}^{2} + \textcolor{red}{10 {x}^{2}} + 5 x + \textcolor{b l u e}{44 x} + \textcolor{\mathmr{and} a n \ge}{48} = 0 - 0 - 0$

$\left(1 + \textcolor{red}{10}\right) {x}^{2} + \left(5 + \textcolor{b l u e}{44}\right) x + 48 = 0$

$11 {x}^{2} + 49 x + 48 = 0$

We can now use the quadratic equation to solve this problem:

For $\textcolor{red}{a} {x}^{2} + \textcolor{b l u e}{b} x + \textcolor{g r e e n}{c} = 0$, the values of $x$ which are the solutions to the equation are given by:

$x = \frac{- \textcolor{b l u e}{b} \pm \sqrt{{\textcolor{b l u e}{b}}^{2} - \left(4 \textcolor{red}{a} \textcolor{g r e e n}{c}\right)}}{2 \cdot \textcolor{red}{a}}$

Substituting:

$\textcolor{red}{11}$ for $\textcolor{red}{a}$

$\textcolor{b l u e}{49}$ for $\textcolor{b l u e}{b}$

$\textcolor{g r e e n}{48}$ for $\textcolor{g r e e n}{c}$ gives:

x = (-color(blue)(49) +- sqrt(color(blue)(49)^2 - (4 * color(red)(11) * color(green)(48))))/(2 * color(red)(11)

$x = \frac{- \textcolor{b l u e}{49} \pm \sqrt{2401 - 2112}}{22}$

$x = \frac{- \textcolor{b l u e}{49} \pm \sqrt{289}}{22}$

$x = \frac{- \textcolor{b l u e}{49} - 17}{22}$ and $x = \frac{- \textcolor{b l u e}{49} + 17}{22}$

$x = - \frac{66}{22}$ and $x = - \frac{32}{22}$

$x = - 3$ and $x = - \frac{16}{11}$