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
Hint : This works exactly the same as the first four problems even though the \(x\)’s are in the denominator. The only difference here is that the \(x\)’s will be in the denominator of our substitution.
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 : \[\require{bbox} \bbox[2pt,border:1px solid black]{{x = - \frac{2}{3}\,\,{\mbox{and }}x = - \frac{1}{7}}}\] .