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### Manipulation of (Imaginary?) Roots

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Date: 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
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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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Associated Topics:
High School Basic Algebra
High School Imaginary/Complex Numbers

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