Section 2.9 : Equations Reducible to Quadratic in Form
1. Solve the following equation.
\[{x^6} - 9{x^3} + 8 = 0\]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 \(6 = 2\left( 3 \right)\) and so we can use the following substitution to reduce the equation to a quadratic equation.
\[u = {x^3}\hspace{0.25in}\hspace{0.25in}{u^2} = {\left( {{x^3}} \right)^2} = {x^6}\] Show Step 2Using this substitution the equation becomes,
\[\begin{align*}{u^2} - 9u + 8 & = 0\\ \left( {u - 1} \right)\left( {u - 8} \right) & = 0\end{align*}\]We can easily see that the solution to this equation is : \(u = 1\) and \(u = 8\) .
Show Step 3Now all we need to do is use our substitution from the first step to determine the solution to the original equation.
\[u = 1:\hspace{0.25in}{x^3} = 1\hspace{0.25in} \Rightarrow \hspace{0.25in}x = {\left( 1 \right)^{\frac{1}{3}}} = 1\] \[u = 8:\hspace{0.25in}{x^3} = 8\hspace{0.25in} \Rightarrow \hspace{0.25in}x = {\left( 8 \right)^{\frac{1}{3}}} = 2\]Therefore the two solutions to the original equation are : \(\require{bbox} \bbox[2pt,border:1px solid black]{{x = 1\,\,{\mbox{and }}x = 2}}\) .