Surd TrickDate: 02/14/2012 at 21:07:59 From: Praveen Subject: Surd Find the value of SQRT[49 - 20 SQRT(6)] In case the meaning of these nested radicals is not clear, please see this image: http://www.imagebam.com/image/c7505a174856941 I cannot reduce SQRT(49) to 7, since something else gets subtracted from it first. So I tried letting ... x = SQRT[49 - 20 SQRT(6)] ... and then squaring both the sides, but I still couldn't get the answer. Next, I tried to move the "20" inside the SQRT(6), which also was no use. Please help! Show me how to solve this. Thanks a lot. Date: 02/16/2012 at 19:50:36 From: Doctor Vogler Subject: Re: Surd Hi Praveen, Thanks for writing to Dr. Math. Here is a trick for taking square roots of quadratic surds. You have something of the form a - sqrt(b). (In your case, a = 49, and b = 6*20^2.) We assume that b is not a square. You want to find a square root of the form ... n - sqrt(d) ... or possibly ... sqrt(c) - sqrt(d). So you are looking for rational solutions (in c and d, given a and b) that solve this equation: sqrt(c) - sqrt(d) = sqrt(a - sqrt(b)) Equivalently: (sqrt(c) - sqrt(d))^2 = a - sqrt(b) Well, the last equation simplifies to c + d - 2*sqrt(c*d) = a - sqrt(b). Since b is not a square, and a, b, c, and d are all rational numbers, the only way there could possibly be a solution is if c + d = a and 2*sqrt(c*d) = sqrt(b). In other words, c + d = a 4*c*d = b. Now here's the big leap: It turns out that if this happens, then a^2 - b = (c - d)^2. So if you check that a^2 - b is not a perfect square, then the square root of your quadratic surd is not quadratic (or biquadratic) but rather quartic. In this case, you can't simplify the square root. But if this ... a^2 - b = r^2 ... is the square of some rational number, then you can declare c = (a + r)/2 d = (a - r)/2 Now, you will get both ... sqrt(c) - sqrt(d) = sqrt(a - sqrt(b)) ... and ... sqrt(c) + sqrt(d) = sqrt(a + sqrt(b)). Of course, if c or d is itself a perfect square (as happens rather often), then you can simplify further. If you have any questions about this or need more help, please write back and show me what you have been able to do, and I will try to offer further suggestions. - Doctor Vogler, The Math Forum http://mathforum.org/dr.math/ |
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