Simplifying RadicalsDate: 06/18/2001 at 21:19:03 From: Alec Colvin Subject: Simplifying radicals I was playing around with radical expression for the trig functions when I noticed the following: sin(15) = sqrt(2-sqrt(3))/2 = (sqrt(6)-sqrt(2))/4 How do you manipulate sqrt(2-sqrt(3))/2 to obtain (sqrt(6)-sqrt(2))/4? Thanks. Date: 06/19/2001 at 10:15:07 From: Doctor Peterson Subject: Re: Simplifying radicals Hi, Alec. Not all expressions of this form can be simplified; but if one can, we would like the square root to have the form sqrt(a) + sqrt(b) or sqrt(a) - sqrt(b) where a and b are positive rational numbers, and a > b to give the positive square root. If we assume that sqrt(2-sqrt(3)) has this form, then this must be true: (sqrt(a) +- sqrt(b))^2 = 2 - sqrt(3) Expanding, we get a + b +- 2 sqrt(ab) = 2 - sqrt(3) Equating the rational and irrational parts of this equation, a + b = 2 +-2 sqrt(ab) = -sqrt(3) The second equation here tells us that we will have to choose the negative sign in order to have a solution, since square roots have to be positive. If we now square this equation, we have a + b = 2 4ab = 3 Solving these equations by substituting in the second, we get 4a(2 - a) = 3 4a^2 - 8a - 3 = 0 8 +- sqrt(64 + 48) a = ------------------ = 1/2 or 3/2 8 and b = 2 - a = 3/2 or 1/2. One choice here will give the positive root, the other the negative root. We take a > b, so that a = 3/2 and b = 1/2. Consequently, our solution is sqrt(2 - sqrt(3)) = sqrt(3/2) - sqrt(1/2) = (sqrt(6) - sqrt(2))/2 Looking for information on this technique, which I didn't recall being taught, I found this page, which together with some rather advanced techniques includes a quote from an old algebra text about the method: Fun with algebraic number theory - Dave Rusin http://www.math.niu.edu/~rusin/known-math/99/nested_rad Here's a page I just found in our archives about another approach: Un-nesting Radicals http://mathforum.org/dr.math/problems/tran.03.16.01.html - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/