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

Solve for x:

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

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- Aug 18th 2012, 04:20 PMnpierceQuadratic Equation
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 PMjgv115Re: Quadratic Equation
Let $\displaystyle u=x^\frac{1}{6}$

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

Go from there - Aug 18th 2012, 05:38 PMnpierceRe: Quadratic Equation
Can you explain how you got u^2 ? thank you

- Aug 18th 2012, 06:25 PMjgv115Re: Quadratic Equation
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 PMnpierceRe: Quadratic Equation
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 PMWilmerRe: Quadratic Equation
If a^p = b, then a = b^(1/p)

- Aug 18th 2012, 08:34 PMjgv115Re: Quadratic Equation
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 AMnpierceRe: Quadratic Equation
- Aug 19th 2012, 11:48 AMWilmerRe: Quadratic Equation