Chapter 1.6, Problem 30E

To determine

To calculate: The solution of the equation, $3x^{\frac{1}{3}}+2x^{\frac{2}{3}}=5$, and check the solution.

The solution of the given equation, $3x^{\frac{1}{3}}+2x^{\frac{2}{3}}=5$ is $x=\frac{-125}{8},1$.

Calculation:

Consider the provided equation,

$3x^{\frac{1}{3}}+2x^{\frac{2}{3}}=5$

Put $x^{\frac{1}{3}}=u$ and convert this equation into quadratic form.

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

Now factorize the above equation.

$\left(2u+5\right)\left(u-1\right)=0$

Put the first factor equal to zero, then replace $u$ with $x^{\frac{1}{3}}$, and solve for $x$.

$\begin{array}{c}2u+5=0\\ u=\frac{-5}{2}\\ x^{\frac{1}{3}}=\frac{-5}{2}\\ x=\frac{-125}{8}\end{array}$

Put the second factor equal to zero, then replace $u$ with $x^{\frac{1}{3}}$, and solve for $x$.

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

Check,

Put $x=\frac{-125}{8}$ in the given equation, $3x^{\frac{1}{3}}+2x^{\frac{2}{3}}=5$

$\begin{array}{r}3{\left(\frac{-125}{8}\right)}^{\frac{1}{3}}+2{\left(\frac{-125}{8}\right)}^{\frac{2}{3}}\stackrel{?}{=}5\\ 3\left(\frac{-5}{2}\right)+2{\left(\frac{-5}{2}\right)}^2\stackrel{?}{=}5\\ \frac{-15}{2}+\frac{25}{2}\stackrel{?}{=}5\\ 5=5\end{array}$

Which is true.

Put $x=1$ in the given equation,$3x^{\frac{1}{3}}+2x^{\frac{2}{3}}=5$.

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

Which is true.

Hence, the solution of the given equation is $x=\frac{-125}{8},1$.