Manipulation of (Imaginary?) RootsDate: 8/18/96 at 19:46:41 From: Anonymous Subject: Manipulation of (Imaginary?) Roots I would appreciate it if you could point out the mistake I'm making. Let r,s, and t be the roots of x^3-6x^2+5x-7=0. Find 1/r^2+1/s^2+1/t^2. Simplifying the fractions into one, I get [(st)^2+(rt)^2+(sr)^2]/(rst)^2 Let the numerator be x. Expand (rs+st+rt)^2 = x+2rst(r+s+t) Using the coefficients from the original equation I get 25 = x+2*7*6 or x = -59 and the desired fraction is -59/49. The problem is that everything is squared in the desired fractions, so they cannot be negative. Two of the roots are imaginary, but I don't think that should affect the result. Thanks, Michael Date: 8/19/96 at 4:34:42 From: Doctor Pete Subject: Re: Manipulation of (Imaginary?) Roots The answer you obtained is indeed correct. In fact, it is precisely because two of the roots are imaginary that the above result is possible! To see why this is so, take a look at the original expression, 1/r^2 + 1/s^2 + 1/t^2 . If, say, r and s were imaginary, r^2 and s^2 would be real and *negative*. Thus 1/r^2 and 1/s^2 would be negative, and one could easily expect the entire sum to be negative as well. In this case, the approximate numerical values of the three roots are {5.30633, 0.346833 + 1.09494 I, 0.346833 - 1.09494 I} Since the squares of complex conjugates are also complex conjugates, it is clear that the imaginary component vanishes when the sum of the reciprocals is taken. Since the real root is relatively large, the reciprocal of its square will be very small - small enough that the net result is negative. -Doctor Pete, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
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