Answer
$$x = \frac{97\frac{+}{} \sqrt {193}}{72}$$
Work Step by Step
$$x^{\frac{1}{2}} - 6x^{\frac{1}{4}} + 8 = 0$$
We can re-write this equation in the following manner: $(x^{\frac{1}{2}}) - 6(x^{\frac{1}{2}})^{2} + 8 = 0$. Substituting the term $x^{\frac{1}{2}}$for an arbitrary variable $y$, we can now express the equation as:
$$y - 6y^{2} + 8 = 0$$
And solve using the quadratic formula $\frac{-b\frac{+}{} \sqrt {b^{2} - 4ac}}{2a}$
$$ y = \frac{-(1)\frac{+}{} \sqrt {(1)^{2} - 4(-6)(8)}}{2(-6)}$$
$$ y = \frac{-1\frac{+}{} \sqrt {1 + 192}}{-12}$$
$$ y = \frac{-1\frac{+}{} \sqrt {193}}{-12}$$
Substituting $y$ to its original value:
$$(x^{\frac{1}{2}}) = \frac{-1\frac{+}{} \sqrt {193}}{-12}$$
$$(x^{\frac{1}{2}})^{2} = (\frac{-1\frac{+}{} \sqrt {193}}{-12})^{2}$$
$$x = \frac{1 + 193\frac{+}{} 2\sqrt {193}}{144}$$
$$x = \frac{194\frac{+}{} 2\sqrt {193}}{144}$$
$$x = \frac{97\frac{+}{} \sqrt {193}}{72}$$