Question 1209723: Let p, q, r, and s be the roots of g(x) = 3x^4 - 8x^3 + 5x^2 + 2x - 17 - 2x^4 + 10x^3 + 11x^2 + 18x - 14.
Compute p^2 + q^2 + r^2 + s^2.
Answer by ikleyn(52879)   (Show Source):
You can put this solution on YOUR website! .
Let p, q, r, and s be the roots of g(x) = 3x^4 - 8x^3 + 5x^2 + 2x - 17 - 2x^4 + 10x^3 + 11x^2 + 18x - 14.
Compute p^2 + q^2 + r^2 + s^2.
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Solution
Reduce the polynomial to the standard form, combining like terms
3x^4 - 8x^3 + 5x^2 + 2x - 17 - 2x^4 + 10x^3 + 11x^2 + 18x - 14 = x^4 + 2x^3 + 16x^2 + 20x - 31.
Notice that
p^2 + q^2 + r^2 + s^2 = (p + q + r + s)^2 - 2*(pq + pr +ps + qr + qs + rs) (standard decomposition for the square of a sum)
Also notice that the leading coefficient of the polynomial standard form at x^4 is 1.
Due to Vieta's theorem
pq + pr + ps + qr + qs + rs = 16 (the sum of in-pairs products of the roots is the coefficient at x^2)
p + q + r + s = -2 (the sum of the roots is the coefficient at x^3 with the opposite sign).
Therefore,
p^2 + q^2 + r^2 + s^2 = (-2)^2 - 2*16 = 4 - 32 = -28.
ANSWER. p^2 + q^2 + r^2 + s^2 = -28.
(by the way, since the sum of squares is negative, it means that some roots are complex numbers).
Solved.
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