Chapter 1.6, Problem 46E

To determine

To calculate: The solution of the equation ${\left(x+2\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=9$.

The solutions of the equation ${\left(x+2\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=9$ are 25 and $-29$.

Calculation:

Consider the provided equation,

${\left(x+2\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=9$

Rewrite in radical form.

$\sqrt[3]{{\left(x+2\right)}^2}=9$

Cube each side and solve,

$\begin{array}{c}{\left(x+2\right)}^2={\left(9\right)}^3\\ {\left(x+2\right)}^2=729\\ x+2=\sqrt{729}\\ x+2=\pm 27\end{array}$

This becomes absolute value equation.

$x+2=\left|27\right|$

In order to solve this equation, the expression inside the absolute value bars can be positive or negative. Therefore, solve the expression two times.

First, solve by using the positive sign.

$\begin{array}{c}x+2=27\\ x=25\end{array}$

Now, solve by using the negative sign.

$\begin{array}{c}x+2=-27\\ x=-29\end{array}$

Check:

Put $x=25$ in the equation ${\left(x+2\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=9$.

$\begin{array}{r}{\left(25+2\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\stackrel{?}{=}9\\ {\left(27\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\stackrel{?}{=}9\\ {\left(3\right)}^2\stackrel{?}{=}9\\ 9=9\end{array}$

Which is true.

Put $x=-29$ in the equation ${\left(x+2\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}=9$.

$\begin{array}{r}{\left(-29+2\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\stackrel{?}{=}9\\ {\left(-27\right)}^{\raisebox{1ex}{$2$}\!\left/ \!\raisebox{-1ex}{$3$}\right.}\stackrel{?}{=}9\\ {\left(-3\right)}^2\stackrel{?}{=}9\\ 9=9\end{array}$

Which is true.

Hence, the solution of given equation is $x=25,-29$.