HollyDay Posted October 22, 2012 Share Posted October 22, 2012 I do not follow this solution: Simplify the following: the 4th root of 3 divided by the 4th root of 5 The solution manual says to multiply the top and bottom by the 4th root of 5 cubed. I don't see why. Why not just multiply the top and bottom by the 4th root of 5? Quote Link to comment Share on other sites More sharing options...

regentrude Posted October 22, 2012 Share Posted October 22, 2012 They want to you rationalize the denominator. The 4th root of 3 divided by the 4th root of 5 is of course the 4th root of (3/5). However, one often wants to have a rational denominator. To rationalize the denominator, you need to multiply both numerator and denominator by 4th root of 5 cubed (i.e. 5^(3/4))so that your denominator ends up being 5, a rational number: (3/5)^(1/4)= (3/5)^(1/4) * (5/5/)^(3/4) = [3^(1/4) * 5^(3/4)] /5 Quote Link to comment Share on other sites More sharing options...

iwantsprinkles Posted October 22, 2012 Share Posted October 22, 2012 Exactly. Here is a visual example http://www.helpalgebra.com/articles/rationalizedenominator.htm Quote Link to comment Share on other sites More sharing options...

TranquilMind Posted October 22, 2012 Share Posted October 22, 2012 They want to you rationalize the denominator.The 4th root of 3 divided by the 4th root of 5 is of course the 4th root of (3/5). However, one often wants to have a rational denominator. To rationalize the denominator, you need to multiply both numerator and denominator by 4th root of 5 cubed (i.e. 5^(3/4))so that your denominator ends up being 5, a rational number: (3/5)^(1/4)= (3/5)^(1/4) * (5/5/)^(3/4) = [3^(1/4) * 5^(3/4)] /5 No wonder your 15 year old is in Calculus. Quote Link to comment Share on other sites More sharing options...

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