I have been informed that on March 7th from 6:00am to 6:00pm Central Time Lamar University will be doing some maintenance to replace a faulty UPS component and to do this they will be completely powering down their data center.
Unfortunately, this means that the site will be down during this time. I apologize for any inconvenience this might cause.
Paul
February 18, 2026
Section 2.9 : Equations Reducible to Quadratic in Form
5. Solve the following equation.
\[\frac{2}{{{x^2}}} + \frac{{17}}{x} + 21 = 0\]Show All Steps Hide All Steps
First let’s notice that \(2 = 2\left( 1 \right)\) and so we can use the following substitution to reduce the equation to a quadratic equation.
\[u = \frac{1}{x}\hspace{0.25in}\hspace{0.25in}{u^2} = {\left( {\frac{1}{x}} \right)^2} = \frac{{{1^2}}}{{{x^2}}} = \frac{1}{{{x^2}}}\] Show Step 2Using this substitution the equation becomes,
\[\begin{align*}2{u^2} + 17u + 21 & = 0\\ \left( {2u + 3} \right)\left( {u + 7} \right) & = 0\end{align*}\]We can easily see that the solution to this equation is : \(u = - \frac{3}{2}\) and \(u = - 7\) .
Show Step 3Now all we need to do is use our substitution from the first step to determine the solution to the original equation.
\[u = - \frac{3}{2}:\hspace{0.25in}\frac{1}{x} = - \frac{3}{2}\hspace{0.25in} \Rightarrow \hspace{0.25in}x = \frac{1}{{ - \frac{3}{2}}} = - \frac{2}{3}\] \[u = - 7:\hspace{0.25in}\frac{1}{x} = - 7\hspace{0.25in} \Rightarrow \hspace{0.25in}x = \frac{1}{{ - 7}} = - \frac{1}{7}\]Therefore the two solutions to the original equation are : \[\require{bbox} \bbox[2pt,border:1px solid black]{{x = - \frac{2}{3}\,\,{\mbox{and }}x = - \frac{1}{7}}}\] .