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August 27th, 2017, 01:06 AM   #1
App
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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.
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Last edited by skipjack; August 27th, 2017 at 02:50 AM.
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August 27th, 2017, 01:20 AM   #2
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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.
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August 27th, 2017, 01:28 AM   #3
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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.
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August 27th, 2017, 09:46 AM   #4
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Quote:
Originally Posted by App View Post
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?
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August 27th, 2017, 09:56 AM   #5
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$x^{4/3} - 5x^{2/3} - 6 = 0$

$(x^{2/3} - 2)(x^{2/3} - 3) = 0$


$x^{2/3} = 2$

$\left(x^{2/3}\right)^{3/2} = 2^{3/2}$

$x = \sqrt{2^3} = \sqrt{2^2 \cdot 2} = 2\sqrt{2}$


$x^{2/3} = 3$

$\left(x^{2/3}\right)^{3/2} = 3^{3/2}$

$x = \sqrt{3^3} = \sqrt{3^2 \cdot 3} = 3\sqrt{3}$
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August 27th, 2017, 11:13 AM   #6
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$n^{3/2} = n^{1 + 1/2} = n^1n^{1/2} = n√n$
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September 1st, 2017, 05:18 AM   #7
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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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