Simplifying Square Root within Square Root without CalculatorDate: 07/05/2004 at 08:37:07 From: Ahmad Subject: sqrt under sqrt I need to find sqrt[5 + sqrt(2)] without using a calculator. Is there a general formula for solving a problem like this? Date: 07/05/2004 at 10:25:19 From: Doctor Vogler Subject: Re: sqrt under sqrt Hi Ahmad, Thanks for writing to Dr Math. There is a nice little technique that can unravel the square root of a number like yours. It doesn't work with higher roots (like cube roots), but it works well for square roots. It works like this: To find sqrt(a + b*sqrt(c)), first calculate a^2 - b^2*c. If it is not a square, then you can't simplify it. If a^2 - b^2*c = q^2, then sqrt(a + |b|*sqrt(c)) = sqrt((a+q)/2) + sqrt((a-q)/2) sqrt(a - |b|*sqrt(c)) = sqrt((a+q)/2) - sqrt((a-q)/2). For example, let's try sqrt(8 + sqrt(15)). In this case, a = 8, b = 1, and c = 15. So: a^2 - b^2*c = 8^2 - 1^2*15 = 64 - 15 = 49 Since 49 is a perfect square, q^2 = 49 and q = 7, and we can simplify to sqrt((8+7)/2) + sqrt((8-7)/2) or sqrt(15/2) + sqrt(1/2). We can simplify that answer a bit, and then we'll check it by squaring it to see if it really gives us the quantity we originally were trying to take the square root of. Simplifying, we get: sqrt(15/2) + sqrt(1/2) = sqrt(15)/sqrt(2) + sqrt(1)/sqrt(2) = [sqrt(15) + sqrt(1)] / sqrt(2) = [sqrt(15) + 1] / sqrt(2) Now let's square that simplified answer to check it: {[sqrt(15) + 1] / sqrt(2)}^2 (square the binomial) [15 + 2sqrt(15) + 1] / 2 [16 + 2sqrt(15)] / 2 8 + sqrt(15) We started by trying to find the square root of (8 + sqrt(15)), and our answer does in fact return that value when squared, so we are correct. If you have any questions about this or need more help, please write back, and I will try to offer further suggestions. - Doctor Vogler, 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/