Cube and Square Roots
Date: 1 Mar 1995 19:29:18 -0500 From: Anonymous Subject: Re: math problems Dear Dr. Math, Can you stand some more problems from teachers? Answers in 24 hours will merit you a bonus! (1) What is the sum of the cubes of the roots of 3x^3 + 4x + 2 = 0 ? (Is there a short way to get this without actually finding the roots? (2) Simplify: sqrt(3 - sqrt(5))/ (sqrt(2) + sqrt(7 - 3sqrt(5))) or (3 - (5^.5)^.5)/ (2^.5 + ( 7 - 3*5^.5))^.5 These two are supposed to say the same thing; however, I have trouble seeing the math symbols all written this way, so I'm not sure they do. Thanks, Janet Ramser Clarksville, TN
Date: 1 Mar 1995 23:45:52 -0500 From: Dr. Ken Subject: Re: math problems Well, I'll see what I can do. For the first problem, try this: every polynomial factors into linear factors in the complex numbers. So we can write this polynomial as 3(x-r1)(x-r2)(x-r3). If you multiply this polynomial out, you'll get the following polynomial: -3 r1*r2*r3 + 3(r1*r2 + r1*r3 + r2*r3)x - 3(r1 + r2 + r3)x^2 + 3x^3 So we know that -3(r1*r2*r3) = 2, and so on, in terms of the coefficients of the original polynomial. Now, let's say we cube the 3(r1 + r2 + r3) term. That will give us r1^3 + r2^3 + r3^3 plus some other garbage. We want to get rid of that other garbage, which we can do. Notice that if we subtract off 3(r1*r2 + r1*r3 + r2*r3)(r1*r2*r3) and also 3(r1*r2*r3), we'll get rid of all the garbage! See if this method pans out for you. The second one is quite similar to the one that you sent in a little while ago: the key will be simplifying the Square root of 7 - 3Sqrt(5) and the Square root of 3 - Sqrt(5). Like before, this will involve systems of equations, being 2ab = 3, a^2 + 5b^2 = 7 and 2cd = 1, c^2 + 5d^2 = 3. If you solve these equations, you should unravel the problem. Let me know if this is too vague or if I just don't make sense. -Ken "Dr." Math
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