# Math Help - Quadratic Equation

Not sure where to start on this. It's in chapter covering complete the square. Any help would be great.

Solve for x:

$x^\frac{1}{3}+3x^\frac{1}{6}-28=0$

Let $u=x^\frac{1}{6}$

$u^2+3u-28=0$

Go from there

Can you explain how you got u^2 ? thank you

If $u=x^\frac{1}{6}$

Then $u^2 = (x^\frac{1}{6})^2$

Which equals $x^{2*\frac{1}{6}} = x^\frac{1}{3}$

ok... Here's what i have so far:

$u^2+3u-28=0$

$(u-4)(u+7)=0$

$u-4=0$ or $u+7=0$

$u=4$ $u=-7$
$x^\frac{1}{6}=4$ $x^\frac{1}{6}=-7$

i dont know how to handle the fraction part at the end.... Thank YOu!

If a^p = b, then a = b^(1/p)

Yeah adding to what Wilmer said,

If we had for example $x^\frac{1}{2} = 2$. To make x the subject, we had square both sides: $(x^\frac{1}{2})^2 = 2^2$

Originally Posted by jgv115
Yeah adding to what Wilmer said,

If we had for example $x^\frac{1}{2} = 2$. To make x the subject, we had square both sides: $(x^\frac{1}{2})^2 = 2^2$
Sorry, i'm still at a loss...

So if $x^\frac{1}{6}=4$
then to make x the subject..... i'd $4^6$???

So if $x^\frac{1}{6}=4$
then to make x the subject..... i'd $4^6$???