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
6. Solve the following equation.
\[\frac{1}{x} - \frac{{11}}{{\sqrt x }} + 18 = 0\]Show All Steps Hide All Steps
First let’s notice that \(1 = 2\left( {\frac{1}{2}} \right)\) and recall that \(\sqrt x = {x^{\frac{1}{2}}}\) . So we can use the following substitution to reduce the equation to a quadratic equation.
\[u = \frac{1}{{\sqrt x }}\hspace{0.25in}\hspace{0.25in}{u^2} = {\left( {\frac{1}{{\sqrt x }}} \right)^2} = {\left( {\frac{1}{{{x^{\frac{1}{2}}}}}} \right)^2} = \frac{1}{x}\] Show Step 2Using this substitution the equation becomes,
\[\begin{align*}{u^2} - 11u + 18 & = 0\\ \left( {u - 2} \right)\left( {u - 9} \right) & = 0\end{align*}\]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.
\[u = 2:\hspace{0.25in}\frac{1}{{\sqrt x }} = 2\hspace{0.25in} \Rightarrow \hspace{0.25in}\sqrt x = \frac{1}{2}\hspace{0.25in} \Rightarrow \hspace{0.25in}x = {\left( {\frac{1}{2}} \right)^2} = \frac{1}{4}\] \[u = 9:\hspace{0.25in}\frac{1}{{\sqrt x }} = 9\hspace{0.25in} \Rightarrow \hspace{0.25in}\sqrt x = \frac{1}{9}\hspace{0.25in} \Rightarrow \hspace{0.25in}x = {\left( {\frac{1}{9}} \right)^2} = \frac{1}{{81}}\]Therefore the two solutions to the original equation are : \[\require{bbox} \bbox[2pt,border:1px solid black]{{x = \frac{1}{4}\,\,{\mbox{and }}x = \frac{1}{{81}}}}\] .