How do you factor x^-4 -13x^-2 +36 =0?

May 6, 2015

$x = \pm \frac{1}{2} , \pm \frac{1}{3}$

Multiply by ${x}^{4}$ to give $36 {x}^{4} - 13 {x}^{2} + 1$
Solve using the quadratic formula for ${x}^{2}$ ${x}^{2} = \frac{13 \pm \sqrt{{13}^{2} - 4.36}}{2.36} = \frac{13 \pm \sqrt{169 - 144}}{72} = \frac{13 \pm \sqrt{25}}{72} = \frac{18}{72} , \frac{8}{72} = \frac{1}{4} , \frac{1}{9}$
Taking square roots gives $x = \pm \frac{1}{2} , \pm \frac{1}{3}$

May 6, 2015

Terminology:
You can factor the expression on the left. You can solve the equation. You can solve an equation by factoring.
(But you don't really factor an equation.)

To factor: ${x}^{- 4} - 13 {x}^{- 2} + 36$

Notice that ${x}^{- 4} = {\left({x}^{- 2}\right)}^{2}$ (The variable expression in the first term is the square of the one in the second term.)

So, is we use a new variable we'll have a quadratic expression.

Let $u = {x}^{- 2}$. This makes ${u}^{2} = {x}^{- 4}$, so the expression becomes:

${u}^{2} - 13 u + 36$ which can be factored:

$\left(u - 4\right) \left(u - 9\right)$ Now go back to $x$'s

${x}^{- 4} - 13 {x}^{- 2} + 36 = \left({x}^{- 2} - 4\right) \left({x}^{- 2} - 9\right)$

To solve by factoring: ${x}^{- 4} - 13 {x}^{- 2} + 36 = 0$

Factor as above, so the question becomes:

Solve: $\left({x}^{- 2} - 4\right) \left({x}^{- 2} - 9\right) = 0$

So we need: $\left({x}^{- 2} - 4\right) = 0$ or $\left({x}^{- 2} - 9\right) = 0$

${x}^{- 2} - 4 = 0$ $\textcolor{w h i t e}{\text{sssss}}$ or $\textcolor{w h i t e}{\text{sssss}}$ ${x}^{- 2} - 9 = 0$

${x}^{- 2} = 4$$\textcolor{w h i t e}{\text{ssssssssss}}$ or $\textcolor{w h i t e}{\text{sssss}}$ ${x}^{- 2} = 9$

$\frac{1}{x} ^ 2 = 4$ $\textcolor{w h i t e}{\text{ssssssssss}}$ or $\textcolor{w h i t e}{\text{sssss}}$ $\frac{1}{x} ^ 2 = 9$

$\frac{1}{4} = {x}^{2}$$\textcolor{w h i t e}{\text{sssssssssss}}$ or $\textcolor{w h i t e}{\text{sssss}}$ $\frac{1}{9} = {x}^{2}$

$x = \pm \frac{1}{2}$ $\textcolor{w h i t e}{\text{ssssssss}}$ or $\textcolor{w h i t e}{\text{sssss}}$ $x = \pm \frac{1}{3}$

There are four solutions: $- \frac{1}{2} , \frac{1}{2} , - \frac{1}{3} , \frac{1}{3}$