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 Vieta's derivatives (Posted on 2016-12-02)

f(x)=x^3+x^2-3x+4

Let the zeroes of the function above be a, b and c

Find f'(a)f'(b)f'(c)

 No Solution Yet Submitted by Danish Ahmed Khan No Rating

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 Analytic Solution Comment 2 of 2 |
Using the coefficients in the cubic equation, we know the
values of the following symmetric functions to be:

S1 = a + b + c = -1,      S2 = ab + bc + ca = -3,      S3 = abc = -4

f’(a) f’(b) f’(c)      = (3a2 + 2a – 3)(3b2 + 2b – 3)(3c2 + 2c – 3)

which can be written* in terms of the symmetric functions as:

-27 + 62S3 + 27S32 + 18S1 + 66S1S3 +27S12 - 66S2 + 18S2S3 – 18S1S2 – 27S22

Using the values given above, this gives:

-27 – 248 + 432 – 18 + 264 + 27 + 198 +216 – 54 – 243

= 547

Without Mathematica this reduction to symmetric functions would
be very tedious. Does anyone know a short cut in  this type of problem?

 Posted by Harry on 2016-12-05 11:07:46

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