Not sure where to start on this. It's in chapter covering complete the square. Any help would be great.
Solve for x:
$\displaystyle x^\frac{1}{3}+3x^\frac{1}{6}-28=0$
ok... Here's what i have so far:
$\displaystyle u^2+3u-28=0$
$\displaystyle (u-4)(u+7)=0$
$\displaystyle u-4=0 $ or$\displaystyle u+7=0$
$\displaystyle u=4$ $\displaystyle u=-7$
$\displaystyle x^\frac{1}{6}=4$ $\displaystyle x^\frac{1}{6}=-7$
i dont know how to handle the fraction part at the end.... Thank YOu!