# Quadratic Equation

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• August 18th 2012, 04:20 PM
npierce
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:

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

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

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

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

Which equals $x^{2*\frac{1}{6}} = x^\frac{1}{3}$
• August 18th 2012, 06:39 PM
npierce
Re: Quadratic Equation
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!
• August 18th 2012, 07:02 PM
Wilmer
Re: Quadratic Equation
If a^p = b, then a = b^(1/p)
• August 18th 2012, 08:34 PM
jgv115
Re: Quadratic Equation
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: $(x^\frac{1}{2})^2 = 2^2$
• August 19th 2012, 11:18 AM
npierce
Re: Quadratic Equation
Quote:

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 $4^6$???
• August 19th 2012, 11:48 AM
Wilmer
Re: Quadratic Equation
Quote:

Originally Posted by npierce
Sorry, i'm still at a loss...
So if $x^\frac{1}{6}=4$
then to make x the subject..... i'd $4^6$???

YES! a^p = b^1 : a = b^(1/p) ; so:
x^(1/6) = 4^1
x = 4^(1 / (1/6))
x = 4^6 = 4096