
Algebra PreAlgebra and Basic Algebra Math Forum 
 LinkBack  Thread Tools  Display Modes 
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: 1,597 Thanks: 546 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: 2,755 Thanks: 1405 
$x^{4/3}  5x^{2/3}  6 = 0$ $(x^{2/3}  2)(x^{2/3}  3) = 0$ $x^{2/3} = 2$ $\left(x^{2/3}\right)^{3/2} = 2^{3/2}$ $x = \sqrt{2^3} = \sqrt{2^2 \cdot 2} = 2\sqrt{2}$ $x^{2/3} = 3$ $\left(x^{2/3}\right)^{3/2} = 3^{3/2}$ $x = \sqrt{3^3} = \sqrt{3^2 \cdot 3} = 3\sqrt{3}$ 
August 27th, 2017, 11:13 AM  #6 
Global Moderator Joined: Dec 2006 Posts: 19,054 Thanks: 1618 
$n^{3/2} = n^{1 + 1/2} = n^1n^{1/2} = n√n$

September 1st, 2017, 05:18 AM  #7 
Math Team Joined: Jan 2015 From: Alabama Posts: 3,191 Thanks: 871 
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 $\displaystyle x= \sqrt{2^3}= 2\sqrt{2}$. 

Tags 
equation, factorial, fractional, powers 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
Determining n from the given factorial equation  mared  Algebra  11  October 2nd, 2014 01:46 AM 
An equation with the factorial  Dacu  Number Theory  5  May 25th, 2013 05:54 AM 
Diophantine equation : factorial and powers  momo  Number Theory  8  May 8th, 2009 12:21 PM 
equation with powers  damgam  Algebra  1  December 19th, 2008 03:17 PM 
An equation with the factorial  Dacu  Algebra  2  December 31st, 1969 04:00 PM 