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August 9th, 2018, 11:23 AM   #1
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nth roots

If (r)^(-1/4) - (r)^(1/4) = 14. Find the value of (r)^(-1/6) + (r)^(1/6)
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August 9th, 2018, 01:05 PM   #2
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r = 19601 - 13860*sqrt(2)
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August 9th, 2018, 01:23 PM   #3
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The answer to this thing is 6.

It can be done via brute force using hyperbolic trig multi-angle formulas but there is probably a very clever way lurking in here that someone like Idea or SlipEternal will see.
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August 9th, 2018, 03:01 PM   #4
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Quote:
Originally Posted by fahad nasir View Post
If (r)^(-1/4) - (r)^(1/4) = 14. Find the value of (r)^(-1/6) + (r)^(1/6)
I'd try a u-substitution.

$u = r^{1/4} \text { and } 14 = r^{-1/4} - r^{1/4} = u^{-1} - u = \dfrac{1}{u} - u \implies$

$\dfrac{1 - u^2}{u} = 14 \implies u^2 + 14u - 1 = 0 \implies u = \dfrac{-\ 14 \pm \sqrt{14^2 - 4(1)(-\ 1)}}{2} = -\ 7 \pm 5 \sqrt{2}.$

Is that correct?

$\dfrac{1}{-\ 7 - 5 \sqrt{2}} - (-\ 7 - 5\sqrt{2}) = \dfrac{-\ 1}{7 + 5 \sqrt{2}} + 7 + 5 \sqrt{2} = \dfrac{49 + 2 * 7 * 5 \sqrt{2} + 25 * 2 - 1}{7 + 5 \sqrt{2}} =$

$\dfrac{98 + 70\sqrt{2}}{7 + 5 \sqrt{2}} = \dfrac{14(7 + 5 \sqrt{2}}{7 + 5 \sqrt{2}} = 14.$

$\dfrac{1}{-\ 7 + 5 \sqrt{2}} - (-\ 7 + 5\sqrt{2}) = \dfrac{-\ 1}{7 - 5 \sqrt{2}} + 7 - 5 \sqrt{2} = \dfrac{49 - 2 * 7 * 5 \sqrt{2} + 25 * 2 - 1}{7 + 5 \sqrt{2}} =$

$\dfrac{98 - 70\sqrt{2}}{7 - 5 \sqrt{2}} = \dfrac{14(7 - 5 \sqrt{2}}{7 + 5 \sqrt{2}} = 14.$

Looks OK.

$v = r^{1/6} \text { and } u = r^{1/4} \implies v^6 = r = u^4 \implies v = (u^4)^{1/6} = u^{2/3}.$

$\therefore r^{-1/6} - r^{1/6} = v^{-1} - v = \dfrac{1}{v} - v = \dfrac{1 - v^2}{v} =$

$\dfrac{1 - (u^{2/3})^2}{u^{2/3}} = \dfrac{1 - \sqrt[3]{u^4}}{\sqrt[3]{u^2}}.$

$u = -\ 7 \pm 5 \sqrt{2} \implies u^2 = 99 \pm 70 \sqrt{2} \implies u^4 = 19601 \pm 13860 \sqrt{2}.$

$\therefore r^{-1/6} - r^{1/6} = \dfrac{1 - \sqrt[3]{19601 \pm 13860 \sqrt{2}}}{\sqrt[3]{99 \pm 70 \sqrt{2}}}.$

Please check this because it was very ugly.
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August 9th, 2018, 04:34 PM   #5
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Math Focus: Calculus/ODEs
Let:

$\displaystyle u=r^{\Large\frac{1}{12}}$

And so:

$\displaystyle u^{-3}-u^{3}=14$

Let:

$\displaystyle S=u^{-1}-u^{1}$

Thus:

$\displaystyle S^2=u^{-2}-2+u^{2}$

And:

$\displaystyle S^3=u^{-3}-3u^{-1}+3u-u^{3}$

This implies:

$\displaystyle S^3+3S-14=0$

Or:

$\displaystyle (S-2)\left(S^2+2S+7\right)=0$

Assuming $\displaystyle S$ is real, we then find:

$\displaystyle S=2$

Thus:

$\displaystyle S^2+2=u^{-2}+u^{2}=r^{-\Large\frac{1}{6}}+r^{\Large\frac{1}{6}}=6$
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August 9th, 2018, 10:15 PM   #6
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Quote:
Originally Posted by JeffM1 View Post
$r^{-1/6} - r^{1/6} =$ . . .
Please check this
It's $r^{1/6} + r^{1/6}$ that is to be calculated.

$r = (\pm\sqrt2 - 1)^{12}$, from which the answer of 6 is easily obtained.
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