Chapter 1.6, Problem 29E

To determine

To calculate: The solution of the equation, $9t^{\frac{2}{3}}+24t^{\frac{1}{3}}=-16$ and check the solution.

The solution of the given equation, $9t^{\frac{2}{3}}+24t^{\frac{1}{3}}=-16$ is $t=\frac{-64}{27}$.

Calculation:

Consider the provided equation,

$9t^{\frac{2}{3}}+24t^{\frac{1}{3}}=-16$

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

$\begin{array}{c}9u^2+24u=-16\\ 9u^2+24u+16=0\end{array}$

Now factorize the above equation.

$\begin{array}{c}{\left(3u+4\right)}^2=0\\ \left(3u+4\right)\left(3u+4\right)=0\end{array}$

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

$\begin{array}{c}3u+4=0\\ u=\frac{-4}{3}\\ t^{\frac{1}{3}}=\frac{-4}{3}\\ t={\left(\frac{-4}{3}\right)}^3\end{array}$

Hence, $t=\frac{-64}{27}$.

Check,

Put $t=\frac{-64}{27}$ in the given equation, $9t^{\frac{2}{3}}+24t^{\frac{1}{3}}=-16$.

$\begin{array}{r}9{\left(\frac{-64}{27}\right)}^{\frac{2}{3}}+24{\left(\frac{-64}{27}\right)}^{\frac{2}{3}}\stackrel{?}{=}-16\\ 9{\left(\frac{-4}{3}\right)}^2+24\left(\frac{-4}{3}\right)\stackrel{?}{=}-16\\ \left(\frac{48}{3}\right)+\left(-32\right)\stackrel{?}{=}-16\\ -16=-16\end{array}$

Which is true.

Hence, the solution of the given equation is $t=\frac{-64}{27}$.