• Aug 18th 2012, 04:20 PM
npierce
Not sure where to start on this. It's in chapter covering complete the square. Any help would be great.

Solve for x:

$\displaystyle x^\frac{1}{3}+3x^\frac{1}{6}-28=0$
• Aug 18th 2012, 05:22 PM
jgv115
Let $\displaystyle u=x^\frac{1}{6}$

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

Go from there
• Aug 18th 2012, 05:38 PM
npierce
Can you explain how you got u^2 ? thank you
• Aug 18th 2012, 06:25 PM
jgv115
If $\displaystyle u=x^\frac{1}{6}$

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

Which equals $\displaystyle x^{2*\frac{1}{6}} = x^\frac{1}{3}$
• Aug 18th 2012, 06:39 PM
npierce
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!
• Aug 18th 2012, 07:02 PM
Wilmer
If a^p = b, then a = b^(1/p)
• Aug 18th 2012, 08:34 PM
jgv115
Yeah adding to what Wilmer said,

If we had for example $\displaystyle x^\frac{1}{2} = 2$. To make x the subject, we had square both sides: $\displaystyle (x^\frac{1}{2})^2 = 2^2$
• Aug 19th 2012, 11:18 AM
npierce
Quote:

Originally Posted by jgv115
Yeah adding to what Wilmer said,

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

Sorry, i'm still at a loss...

So if $\displaystyle x^\frac{1}{6}=4$
then to make x the subject..... i'd $\displaystyle 4^6$???
• Aug 19th 2012, 11:48 AM
Wilmer
So if $\displaystyle x^\frac{1}{6}=4$
then to make x the subject..... i'd $\displaystyle 4^6$???