Chapter 1.6, Problem 26E

To determine

To calculate: The solution of the equation $6{\left(\frac{x}{x+1}\right)}^2+5\left(\frac{x}{x+1}\right)-6=0$.

The solutions of the equation $6{\left(\frac{x}{x+1}\right)}^2+5\left(\frac{x}{x+1}\right)-6=0$ are 2 and $\frac{-3}{5}$.

Calculation:

Consider the provided equation,

$6{\left(\frac{x}{x+1}\right)}^2+5\left(\frac{x}{x+1}\right)-6=0$

Put $\frac{x}{x+1}=u$ and convert this equation into quadratic form.

$6u^2+5u-6=0$

Now, factorize the above equation.

$\left(3u-2\right)\left(2u+3\right)=0$

Put the first factor equal to zero, then replace $u$ with $\frac{x}{x+1}$, and solve for $x$.

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

Put the second factor equal to zero, then replace $u$ with $\frac{x}{x+1}$, and solve for $x$.

$\begin{array}{c}2u+3=0\\ u=\frac{-3}{2}\\ \frac{x}{x+1}=\frac{-3}{2}\\ x=\frac{-3}{5}\end{array}$

Check:

Put $x=2,\frac{-3}{5}$ in the given equation.

First, put $x=2$ in the equation $6{\left(\frac{x}{x+1}\right)}^2+5\left(\frac{x}{x+1}\right)-6=0$ as,

$\begin{array}{r}6{\left(\frac{2}{2+1}\right)}^2+5\left(\frac{2}{2+1}\right)-6\stackrel{?}{=}0\\ 6\left(\frac{4}{9}\right)+5\left(\frac{2}{3}\right)-6\stackrel{?}{=}0\\ 6-6\stackrel{?}{=}0\\ 0=0\end{array}$

Which is true.

Now, put $x=\frac{-3}{5}$ in the equation $6{\left(\frac{x}{x+1}\right)}^2+5\left(\frac{x}{x+1}\right)-6=0$ as,

$\begin{array}{r}6{\left(\frac{\left(\frac{-3}{5}\right)}{\frac{-3}{5}+1}\right)}^2+5\left(\frac{\frac{-3}{5}}{\frac{-3}{5}+1}\right)-6\stackrel{?}{=}0\\ 6{\left(\frac{-3}{2}\right)}^2+5\left(\frac{-3}{2}\right)-6\stackrel{?}{=}0\\ 6-6\stackrel{?}{=}0\\ 0=0\end{array}$

Which is true.

Hence, the solutions of given equation are 2 and $\frac{-3}{5}$.