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 August 9th, 2018, 10:23 AM #1 Member   Joined: Oct 2012 Posts: 78 Thanks: 0 nth roots If (r)^(-1/4) - (r)^(1/4) = 14. Find the value of (r)^(-1/6) + (r)^(1/6) August 9th, 2018, 12:05 PM #2 Math Team   Joined: Oct 2011 From: Ottawa Ontario, Canada Posts: 14,597 Thanks: 1038 r = 19601 - 13860*sqrt(2) Thanks from topsquark August 9th, 2018, 12:23 PM #3 Senior Member   Joined: Sep 2015 From: USA Posts: 2,529 Thanks: 1389 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. Thanks from topsquark August 9th, 2018, 02:01 PM   #4
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Quote:
 Originally Posted by fahad nasir 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.

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

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

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

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

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

Please check this because it was very ugly. August 9th, 2018, 03:34 PM #5 Senior Member   Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 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: $\displaystyle S^2+2=u^{-2}+u^{2}=r^{-\Large\frac{1}{6}}+r^{\Large\frac{1}{6}}=6$ Thanks from greg1313, Maschke, fahad nasir and 2 others August 9th, 2018, 09:15 PM   #6
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Quote:
 Originally Posted by JeffM1 $r^{-1/6} - r^{1/6} =$ . . . 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. Tags nth, roots 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 2 June 23rd, 2014 04:08 PM annakar Calculus 8 December 16th, 2012 04:52 AM e81 Algebra 6 May 22nd, 2011 04:53 PM stuart clark Calculus 5 May 21st, 2011 03:20 PM jamesuminator Algebra 15 December 19th, 2008 08:50 AM

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