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August 27th, 2017, 01:06 AM   #1
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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.
Attached Images IMG_1749.jpg (19.8 KB, 6 views) IMG_1750.jpg (19.8 KB, 5 views)

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: 1,833 Thanks: 649 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 (x-2)(x-3)... 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
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Quote:
 Originally Posted by App Oops, yes, thanks! It's (x-2)(x-3)... will start working on it again and see how things go. P.S. Yes, it's fractional powers.
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?
Attached Images IMG_1751.jpg (19.4 KB, 2 views) August 27th, 2017, 09:56 AM #5 Math Team   Joined: Jul 2011 From: Texas Posts: 3,002 Thanks: 1587 $x^{4/3} - 5x^{2/3} - 6 = 0$ $(x^{2/3} - 2)(x^{2/3} - 3) = 0$ $x^{2/3} = 2$ $\left(x^{2/3}\right)^{3/2} = 2^{3/2}$ $x = \sqrt{2^3} = \sqrt{2^2 \cdot 2} = 2\sqrt{2}$ $x^{2/3} = 3$ $\left(x^{2/3}\right)^{3/2} = 3^{3/2}$ $x = \sqrt{3^3} = \sqrt{3^2 \cdot 3} = 3\sqrt{3}$ August 27th, 2017, 11:13 AM #6 Global Moderator   Joined: Dec 2006 Posts: 20,919 Thanks: 2201 $n^{3/2} = n^{1 + 1/2} = n^1n^{1/2} = n√n$ 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 $\displaystyle x= \sqrt{2^3}= 2\sqrt{2}$. Tags equation, factorial, fractional, powers Thread Tools Show Printable Version Email this Page Display Modes Linear Mode Switch to Hybrid Mode Switch to Threaded Mode Similar Threads Thread Thread Starter Forum Replies Last Post mared Algebra 11 October 2nd, 2014 01:46 AM Dacu Number Theory 5 May 25th, 2013 05:54 AM momo Number Theory 8 May 8th, 2009 12:21 PM damgam Algebra 1 December 19th, 2008 03:17 PM Dacu Algebra 2 December 31st, 1969 04:00 PM

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