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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

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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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