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

Jan 29, 2015

Raise both sides to the 4th power:
${\left({\left(3 - 8 {x}^{2}\right)}^{\frac{1}{4}}\right)}^{4}$ = ${\left(2 x\right)}^{4}$

Simplify:
$3 - 8 {x}^{2} = {2}^{4} \cdot {x}^{4}$
$3 - 8 {x}^{2} = 16 {x}^{4}$
$0 = 16 {x}^{4} + 8 {x}^{2} - 3$
$0 = \left(4 {x}^{2} - 1\right) \left(4 {x}^{2} + 3\right)$

So: $4 {x}^{2} - 1 = 0$ or $4 {x}^{2} + 3 = 0$

$4 {x}^{2} - 1 = 0$ -> $4 {x}^{2} = 1$ -> ${x}^{2} = \frac{1}{4}$ -> $x = \pm \frac{1}{2}$
$4 {x}^{2} + 3 = 0$ -> $4 {x}^{2} = - 3$ -> not a real solution

Now we have to check for extraneous solutions:
$x = \frac{1}{2}$:
Left side: ${\left(3 - 8 \cdot \left(\frac{1}{4}\right)\right)}^{\frac{1}{4}}$ = ${\left(3 - 2\right)}^{\frac{1}{4}} = {1}^{\frac{1}{4}} = 1$
Right side: $2 \cdot \frac{1}{2} = 1$
Left and right side are equal, so this solution works

$x = - \frac{1}{2}$:
Left side: ${\left(3 - 8 \cdot \left(\frac{1}{4}\right)\right)}^{\frac{1}{4}}$ = ${\left(3 - 2\right)}^{\frac{1}{4}} = {1}^{\frac{1}{4}} = 1$
Right side: $2 \cdot - \frac{1}{2} = - 1$
Left and right side are not equal, so this solution is extraneous.

So our answer: $x = \frac{1}{2}$