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Section 2.9 : Equations Reducible to Quadratic in Form
3. Solve the following equation.
\[4{x^{\frac{2}{3}}} + 21{x^{\frac{1}{3}}} + 27 = 0\]Show All Steps Hide All Steps
Hint : Remember to look at the exponents of the first two terms and try to find a substitution that will turn this into a “normal” quadratic equation.
First let’s notice that \(\frac{2}{3} = 2\left( {\frac{1}{3}} \right)\) and so we can use the following substitution to reduce the equation to a quadratic equation.
\[u = {x^{\frac{1}{3}}}\hspace{0.25in}\hspace{0.25in}{u^2} = {\left( {{x^{\frac{1}{3}}}} \right)^2} = {x^{\frac{2}{3}}}\] Show Step 2Using this substitution the equation becomes,
\[\begin{align*}4{u^2} + 21u + 27 & = 0\\ \left( {4u + 9} \right)\left( {u + 3} \right) & = 0\end{align*}\]We can easily see that the solution to this equation is : \(u = - \frac{9}{4}\) and \(u = - 3\) .
Show Step 3Now all we need to do is use our substitution from the first step to determine the solution to the original equation.
\[u = - \frac{9}{4}:\hspace{0.25in}{x^{\frac{1}{3}}} = - \frac{9}{4}\hspace{0.25in} \Rightarrow \hspace{0.25in}x = {\left( { - \frac{9}{4}} \right)^3} = - \frac{{729}}{{64}}\] \[u = - 3:\hspace{0.25in}{x^{\frac{1}{3}}} = - 3\hspace{0.25in} \Rightarrow \hspace{0.25in}x = {\left( { - 3} \right)^3} = - 27\]Therefore the two solutions to the original equation are : \[\require{bbox} \bbox[2pt,border:1px solid black]{{x = - \frac{{729}}{{64}}\,\,{\mbox{and }}x = - 27}}\] .