Question

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

Answer 1

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 `(x-4)(x+2)=0` and that's it I hope I made it clear

Answer 2

`x=+-2" or "x=+-sqrt2i`

Explanation:

`"make the substitution "u=x^2`

`rArru^2-2u-8=0`

`"the factors of - 8 which sum to - 2 are - 4 and + 2"`

`rArr(u-4)(u+2)=0`

`"equate each factor to zero and solve for u"`

`u-4=0rArru=4`

`u+2=0rArru=-2`

`"change u back into terms of x"`

`u=4rArrx^2=4rArrx=+-2`

`u=-2rArrx^2=-2rArrx=+-sqrt2i`

Answer 3

`x=2, -2,`
If x is not explicitly real, `x = isqrt(2), -isqrt(2)`

Explanation:

Substitute `x^2` for `y`:
`y^2-2y-8=0`
This is just to make it look less intimidating to factor
Factor:
`(y+a)(y+b)`
`ab=-8`
`a+b=-2`
`-4` and `2` work for `a` and `b` (found by trial and error)
`(y-4)(y+2)=0`
Substitute back:
`(x^2-4)(x^2+2)=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:

`(x^2+2)=0`
`(x-isqrt(2))(x+isqrt(2))=0`
`x = 2, -2, isqrt(2), -isqrt(2)`