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### Section 2-9 : Equations Reducible to Quadratic in Form

5. Solve the following equation.

$\frac{2}{{{x^2}}} + \frac{{17}}{x} + 21 = 0$

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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.
Start Solution

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 2

Using 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 3

Now 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 : $\require{bbox} \bbox[2pt,border:1px solid black]{{x = - \frac{2}{3}\,\,{\mbox{and }}x = - \frac{1}{7}}}$ .