Chapter 1.6, Problem 25E

To determine

To calculate: The solution of the equation $2{\left(\frac{x}{x+2}\right)}^2-3\left(\frac{x}{x+2}\right)-2=0$.

The solutions of the equation $2{\left(\frac{x}{x+2}\right)}^2-3\left(\frac{x}{x+2}\right)-2=0$ are $\frac{-2}{3}$ and $-4$.

Calculation:

Consider the provided equation,

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

Put $\frac{x}{x+2}=u$ and convert this equation into quadratic form.

$2u^2-3u-2=0$

Now, factorize the above equation.

$\left(2u+1\right)\left(u-2\right)=0$

Put the first factor equal to zero, then replace $u$ with $\frac{x}{x+2}$, and solve for $x$.

$\begin{array}{c}2u+1=0\\ u=\frac{-1}{2}\\ \frac{x}{x+2}=\frac{-1}{2}\\ x=\frac{-2}{3}\end{array}$

Put the second factor equal to zero, then replace $u$ with $\frac{x}{x+2}$, and solve for $x$.

$\begin{array}{c}u-2=0\\ u=2\\ \frac{x}{x+2}=2\\ x=-4\end{array}$

Check:

Put $x=\frac{-2}{3},-4$ in the given equation.

First, put $x=\frac{-2}{3}$ in the equation $2{\left(\frac{x}{x+2}\right)}^2-3\left(\frac{x}{x+2}\right)-2=0$ as,

$\begin{array}{r}2{\left(\frac{\left(\frac{-2}{3}\right)}{\left(\frac{-2}{3}\right)+2}\right)}^2-3\left(\frac{\left(\frac{-2}{3}\right)}{\left(\frac{-2}{3}\right)+2}\right)-2\stackrel{?}{=}0\\ 2{\left(\frac{-1}{2}\right)}^2-3\left(\frac{-1}{2}\right)-2\stackrel{?}{=}0\\ 2-2\stackrel{?}{=}0\\ 0=0\end{array}$

Which is true.

Now, put $x=-4$ in the equation $2{\left(\frac{x}{x+2}\right)}^2-3\left(\frac{x}{x+2}\right)-2=0$ as,

$\begin{array}{r}2{\left(\frac{-4}{-4+2}\right)}^2-3\left(\frac{-4}{-4+2}\right)-2\stackrel{?}{=}0\\ 2{\left(2\right)}^2-3\left(2\right)-2\stackrel{?}{=}0\\ 8-6-2\stackrel{?}{=}0\\ 0=0\end{array}$

Which is true.

Hence, the solutions of the provided equation are $\frac{-2}{3}$ and $-4$.