Question

Solve the equation: `(3-8x^2)^(1/4) = 2x`?

Answer 1

Raise both sides to the 4th power:
`((3-8x^2)^(1/4))^4` = `(2x)^4`

Simplify:
`3-8x^2 = 2^4*x^4`
`3-8x^2 = 16x^4`
`0 = 16x^4+8x^2-3`
`0 = (4x^2 - 1)(4x^2 + 3)`

So: `4x^2-1 = 0` or `4x^2+3=0`

`4x^2-1 = 0` -> `4x^2=1` -> `x^2 = 1/4` -> `x = +- 1/2`
`4x^2+3=0` -> `4x^2=-3` -> not a real solution

Now we have to check for extraneous solutions:
`x=1/2`:
Left side: `(3-8*(1/4))^(1/4)` = `(3-2)^(1/4) = 1^(1/4) = 1`
Right side: `2*1/2 = 1`
Left and right side are equal, so this solution works

`x=-1/2`:
Left side: `(3-8*(1/4))^(1/4)` = `(3-2)^(1/4) = 1^(1/4) = 1`
Right side: `2*-1/2 = -1`
Left and right side are not equal, so this solution is extraneous.

So our answer: `x=1/2`