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August 27th, 2017, 01:06 AM  #1 
Member Joined: Jan 2014 Posts: 55 Thanks: 2  Equation with fractional powers
Hi, I'm a bit scratchy with one of the last part as you can see underlined in purple with W^2/3 = 6 & W^2/3 = 1 ... not sure whether to cube root it somehow... am just stuck. The answer is in red at the bottom. Need clarification on the steps I took, and why I got stuck, and why the answer is in like the one in red... help asap.
Last edited by skipjack; August 27th, 2017 at 02:50 AM. 
August 27th, 2017, 01:20 AM  #2 
Senior Member Joined: Feb 2016 From: Australia Posts: 1,790 Thanks: 629 Math Focus: Yet to find out. 
I assume you mean fractional powers, not factorial powers. Check your factorisation in the first image. Also, it might be clearer to you if you explicitly write what substitution you are making at the beginning. Namely, $w = x^{2/3}$ so that later on you are solving for $x^{2/3}$ and not w. 
August 27th, 2017, 01:28 AM  #3 
Member Joined: Jan 2014 Posts: 55 Thanks: 2 
Oops, yes, thanks! It's (x2)(x3)... will start working on it again and see how things go. P.S. Yes, it's fractional powers. Last edited by skipjack; August 27th, 2017 at 02:55 AM. 
August 27th, 2017, 09:46 AM  #4 
Member Joined: Jan 2014 Posts: 55 Thanks: 2  I'm still clueless at the fractional exponent. This is where I could reach.How do you rationalize it into sq roots like the answer?

August 27th, 2017, 09:56 AM  #5 
Math Team Joined: Jul 2011 From: Texas Posts: 2,878 Thanks: 1502 
$x^{4/3}  5x^{2/3}  6 = 0$ $(x^{2/3}  2)(x^{2/3}  3) = 0$ $x^{2/3} = 2$ $\left(x^{2/3}\right)^{3/2} = 2^{3/2}$ $x = \sqrt{2^3} = \sqrt{2^2 \cdot 2} = 2\sqrt{2}$ $x^{2/3} = 3$ $\left(x^{2/3}\right)^{3/2} = 3^{3/2}$ $x = \sqrt{3^3} = \sqrt{3^2 \cdot 3} = 3\sqrt{3}$ 
August 27th, 2017, 11:13 AM  #6 
Global Moderator Joined: Dec 2006 Posts: 20,464 Thanks: 2038 
$n^{3/2} = n^{1 + 1/2} = n^1n^{1/2} = n√n$

September 1st, 2017, 05:18 AM  #7 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,264 Thanks: 902 
The way you have done it, immediately after writing the original equation, $\displaystyle x^{4/3} 3x^{2/3}+ 6= 0$ you should have written: "Letting $\displaystyle w= x^{2/3}$ the equation becomes" and then have $\displaystyle w^2 w+ 6= (w 3)(w 2)= 0$ so w= 3 and w= 2. Then $\displaystyle w= 3= x^{2/3}$ so $\displaystyle x= 3^{3/2}$ and $\displaystyle w= 2= x^{2/3}$ so $\displaystyle x= 2^{3/2}$. You could also write those as $\displaystyle x= \sqrt{3^3}= 3\sqrt{3}$ and $\displaystyle x= \sqrt{2^3}= 2\sqrt{2}$. 

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equation, factorial, fractional, powers 
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