Complex RadicalsDate: 11/28/2001 at 17:26:12 From: Anna Subject: Complex radicals I understand how you use the square root, but how do you use the cubed root? The problems are: sqrt(24) --------------------------------------- sqrt(2) + sqrt(3) + sqrt(5) 6 -------------------------- 1 - cubedroot(3) 12 ------------------------------------ cubedroot(4) - cubedroot(2) + 1 Thanks a million! Date: 11/29/2001 at 10:29:04 From: Doctor Peterson Subject: Re: Complex radicals Hi, Anna. The basic idea used in this kind of problem is to use a factorization like these: a^2 - b^2 = (a - b)(a + b) a^3 - b^3 = (a - b)(a^2 + ab + b^2) a^3 + b^3 = (a + b)(a^2 - ab + b^2) Check out those last two if you aren't familiar with them. When a and b are square or cube roots, these formulas tell us we can multiply a-b or a+b by a certain polynomial so that the product will involve only the square or cube of a and b, eliminating the radical. You can do the same with even higher powers; just divide a^n - b^n by a - b, for example, to find the other factor. Your first example involves only square roots, but is tricky because there are three of them: sqrt(24) --------------------------- sqrt(2) + sqrt(3) + sqrt(5) All we can do is try something; so let's take a = sqrt(2) + sqrt(3) and b = sqrt(5). If you multiply numerator and denominator by a - b, which is [sqrt(2) + sqrt(3)] - sqrt(5), you will find that the denominator becomes very simple, and it doesn't take much more work to finish rationalizing the denominator. Other choices may work just as well; try them to familiarize yourself with how these things work. The next one uses the cube root: 6 ----------- 1 - cbrt(3) (Note that it is "cube root", not "cubed root," which would suggest you are cubing the square root. I shortened your abbreviation to my own favorite.) Here you can use my second formula with a = 1 and b = cbrt(3), by multiplying numerator and denominator by a^2 + ab + b^2 = 1 + cbrt(3) + cbrt(9). See what happens! Finally, 12 --------------------- cbrt(4) - cbrt(2) + 1 Here we have both complications; it involves cube roots, and there are three terms in the denominator. My first thought was to do the same thing I did for the first problem, taking part of the denominator as a and part as b, and seeing what happened; you should try that for the experience. You may get a surprise! But working with that made me realize that cbrt(4) is the square of cbrt(2), so that if we just let a = cbrt(2), the denominator is a^2 - a + 1. Look familiar? See if you can use that as a clue. These are a little challenging, but once you get past these, you'll be able to handle most of what they can give you. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ Date: 12/04/2001 at 18:28:33 From: Anna Subject: Re: Complex radicals Thank you. I think I get it. |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2013 The Math Forum
http://mathforum.org/dr.math/