August 9th, 2018, 10:23 AM  #1 
Member Joined: Oct 2012 Posts: 66 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: 13,296 Thanks: 934 
r = 19601  13860*sqrt(2)

August 9th, 2018, 12:23 PM  #3 
Senior Member Joined: Sep 2015 From: USA Posts: 2,122 Thanks: 1102 
The answer to this thing is 6. It can be done via brute force using hyperbolic trig multiangle formulas but there is probably a very clever way lurking in here that someone like Idea or SlipEternal will see. 
August 9th, 2018, 02:01 PM  #4  
Senior Member Joined: May 2016 From: USA Posts: 1,148 Thanks: 479  Quote:
$u = r^{1/4} \text { and } 14 = r^{1/4}  r^{1/4} = u^{1}  u = \dfrac{1}{u}  u \implies$ $\dfrac{1  u^2}{u} = 14 \implies u^2 + 14u  1 = 0 \implies u = \dfrac{\ 14 \pm \sqrt{14^2  4(1)(\ 1)}}{2} = \ 7 \pm 5 \sqrt{2}.$ 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.  
August 9th, 2018, 03:34 PM  #5 
Senior Member Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,204 Thanks: 511 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: $\displaystyle S^3=u^{3}3u^{1}+3uu^{3}$ This implies: $\displaystyle S^3+3S14=0$ Or: $\displaystyle (S2)\left(S^2+2S+7\right)=0$ Assuming $\displaystyle S$ is real, we then find: $\displaystyle S=2$ Thus: $\displaystyle S^2+2=u^{2}+u^{2}=r^{\Large\frac{1}{6}}+r^{\Large\frac{1}{6}}=6$ 
August 9th, 2018, 09:15 PM  #6 
Global Moderator Joined: Dec 2006 Posts: 19,712 Thanks: 1805  

Tags 
nth, roots 
Thread Tools  
Display Modes  

Similar Threads  
Thread  Thread Starter  Forum  Replies  Last Post 
roots  mared  Algebra  2  June 23rd, 2014 04:08 PM 
proof: polynomial n roots, its derivative has n1 roots  annakar  Calculus  8  December 16th, 2012 04:52 AM 
roots  e81  Algebra  6  May 22nd, 2011 04:53 PM 
Sum of roots of f(x)  stuart clark  Calculus  5  May 21st, 2011 03:20 PM 
Relation of Imaginary Roots to Real Roots  jamesuminator  Algebra  15  December 19th, 2008 08:50 AM 