# How do you solve x^4 - 2x^2 - 8 = 0?

Mar 30, 2018

you should ask yourself what two numbers do you multiply to get -8 and their summation is -2
and that is -4 and +2 so $\left(x - 4\right) \left(x + 2\right) = 0$ and that's it I hope I made it clear

Mar 30, 2018

$x = \pm 2 \text{ or } x = \pm \sqrt{2} i$

#### Explanation:

$\text{make the substitution } u = {x}^{2}$

$\Rightarrow {u}^{2} - 2 u - 8 = 0$

$\text{the factors of - 8 which sum to - 2 are - 4 and + 2}$

$\Rightarrow \left(u - 4\right) \left(u + 2\right) = 0$

$\text{equate each factor to zero and solve for u}$

$u - 4 = 0 \Rightarrow u = 4$

$u + 2 = 0 \Rightarrow u = - 2$

$\text{change u back into terms of x}$

$u = 4 \Rightarrow {x}^{2} = 4 \Rightarrow x = \pm 2$

$u = - 2 \Rightarrow {x}^{2} = - 2 \Rightarrow x = \pm \sqrt{2} i$

Mar 30, 2018

$x = 2 , - 2 ,$
If x is not explicitly real, $x = i \sqrt{2} , - i \sqrt{2}$

#### Explanation:

Substitute ${x}^{2}$ for $y$:
${y}^{2} - 2 y - 8 = 0$
This is just to make it look less intimidating to factor
Factor:
$\left(y + a\right) \left(y + b\right)$
$a b = - 8$
$a + b = - 2$
$- 4$ and $2$ work for $a$ and $b$ (found by trial and error)
$\left(y - 4\right) \left(y + 2\right) = 0$
Substitute back:
$\left({x}^{2} - 4\right) \left({x}^{2} + 2\right) = 0$
Factor:
(x-2)(x+2)(x^2+2)=0 x=2, -2#
If you weren't taught imaginary numbers yet, skip the rest of this, you don't need it:

$\left({x}^{2} + 2\right) = 0$
$\left(x - i \sqrt{2}\right) \left(x + i \sqrt{2}\right) = 0$
$x = 2 , - 2 , i \sqrt{2} , - i \sqrt{2}$