please explain...Quote:
Originally posted by alex
7/12
greetinx,
alex
somebody?
anybody...
i tried but did not get it
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please explain...Quote:
Originally posted by alex
7/12
greetinx,
alex
somebody?
anybody...
i tried but did not get it
Hi!
For any polynomial ... + c x^2 + b x + a (a and b nonzero), the sum of the reciprocals of the roots is equal to -b/a. It's that simple if you have that kind of black knowledge! If you want to prove it, it is easiest to start at the other end. If you knew the roots of the polynomial, then you could factor it like this:
and then you can start multiplying... (the roots are called a, b, c, ... below)Code:-v-v- / \
| | | x - r | :cool:
| | \ i i /
See what I mean?Code:(x-a) = x + (-a)
(x-a)(x-b) = x^2 + (-a-b)x + (ab)
(x-a)(x-b)(x-c) = x^3 + (-a-b-c)x^2 + (ab+ac+bc)x + (-abc)
(x-a)(x-b)(x-c)(x-d) = x^4 + (-a-b-c-d)x^3 +
+ (ab+ac+ad+bc+bd+cd)x^2 + (-abc-abd-acd-bcd)x + (abcd)
Now you divide the linear term by the constant term... :eek:
greetinx,
alex