Section 2.9 : Equations Reducible to Quadratic in Form
2. Solve the following equation.
\[{x^{ - 4}} - 7{x^{ - 2}} - 18 = 0\]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 \( - 4 = 2\left( { - 2} \right)\) and so we can use the following substitution to reduce the equation to a quadratic equation.
\[u = {x^{ - 2}}\hspace{0.25in}\hspace{0.25in}{u^2} = {\left( {{x^{ - 2}}} \right)^2} = {x^{ - 4}}\] Show Step 2Using this substitution the equation becomes,
\[\begin{align*}{u^2} - 7u - 18 & = 0\\ \left( {u - 9} \right)\left( {u + 2} \right) & = 0\end{align*}\]We can easily see that the solution to this equation is : \(u = - 2\) and \(u = 9\) .
Show Step 3Now all we need to do is use our substitution from the first step to determine the solution to the original equation.
\[u = - 2:\hspace{0.25in}{x^{ - 2}} = \frac{1}{{{x^2}}} = - 2\hspace{0.25in} \Rightarrow \hspace{0.25in}{x^2} = - \frac{1}{2}\hspace{0.25in} \Rightarrow \hspace{0.25in}x = \pm \sqrt { - \frac{1}{2}} = \pm \frac{1}{{\sqrt 2 }}i\] \[u = 9:\hspace{0.25in}{x^{ - 2}} = \frac{1}{{{x^2}}} = 9\hspace{0.25in} \Rightarrow \hspace{0.25in}{x^2} = \frac{1}{9}\hspace{0.25in} \Rightarrow \hspace{0.25in}x = \pm \sqrt {\frac{1}{9}} = \pm \frac{1}{3}\]Therefore the four solutions to the original equation are : \[\require{bbox} \bbox[2pt,border:1px solid black]{{x = \pm \frac{1}{{\sqrt 2 }}i\,\,{\mbox{and }}x = \pm \frac{1}{3}}}\] .