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

Question

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

2 Answers

Impact of this question

3688 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2015-05-06T14:49:10

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

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

Answer 2 · 2015-05-06T15:59:45

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: $x^{- 4} - 13 x^{- 2} + 36$ $x^(-4)-13x^(-2)+36$

Notice that $x^{- 4} = {(x^{- 2})}^{2}$ $x^(-4) = (x^(-2))^2$ (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 $u = x^{- 2}$ $u=x^(-2)$ . This makes $u^{2} = x^{- 4}$ $u^2 = x^(-4)$ , so the expression becomes:

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

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

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

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

Factor as above, so the question becomes:

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

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

$x^{- 2} - 4 = 0$ $x^(-2)-4=0$ $sssss$ $color(white)"sssss"$ or $sssss$ $color(white)"sssss"$ $x^{- 2} - 9 = 0$ $x^(-2)-9 = 0$

$x^{- 2} = 4$ $x^(-2) = 4$ $ssssssssss$ $color(white)"ssssssssss"$ or $sssss$ $color(white)"sssss"$ $x^{- 2} = 9$ $x^(-2) = 9$

$\frac{1}{x^{2}} = 4$ $1/x^2 = 4$ $ssssssssss$ $color(white)"ssssssssss"$ or $sssss$ $color(white)"sssss"$ $\frac{1}{x^{2}} = 9$ $1/x^2 = 9$

$\frac{1}{4} = x^{2}$ $1/4 = x^2$ $sssssssssss$ $color(white)"sssssssssss"$ or $sssss$ $color(white)"sssss"$ $\frac{1}{9} = x^{2}$ $1/9 = x^2$

$x = \pm \frac{1}{2}$ $x = +- 1/2$ $ssssssss$ $color(white)"ssssssss"$ or $sssss$ $color(white)"sssss"$ $x = \pm \frac{1}{3}$ $x= +- 1/3$

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

Science

Math

Humanities

... and beyond

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

2 Answers

Related questions

Impact of this question

Science

Math

Humanities

... and beyond

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

2 Answers

Related questions

Impact of this question

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