Question

va rog frumos, si explicat daca se poate.

Answer 1

$f = X^4-2X^2+1\\ \\ Fie: x^4-2x^2+1 =0 \\ \\ \left\{ \begin{array}{ll} x_1^4-2x_1^2+1=0 \\ x_2^4-2x_2^2+1=0 \\ x_3^4-2x_3^2+1=0\\ x_4^4-2x_4^2+1=0 \end{array} \right \\ ---------- (+) \\ \\ x_1^4+x_2^4+x_3^4+x_4^4 -2(x_1^2+x_2^2+x_3^2+x_4^2)+1\cdot 4= 0 \\ \\ x_1^4+x_2^4+x_3^4+x_4^4 = 2(x_1^2+x_2^2+x_3^2+x_4^2)-4$

$\Big((x_1+x_2)+(x_3+x_4)\Big) ^2 =\\ = (x_1+x_2)^2+2(x_1+x_2)(x_3+x_4) + (x_3+x_4)^2 = \\ =x_1^2+2x_2+x_2^2+2(x_1x_3+x_1x_4+x_2x_3+x_2x_4) + \\ +x_3^2+2x_3x_4+x_4^2 = \\ = x_1^2+x_2^2+x_3^2+x_4^2+2(x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4)$

$\Rightarrow x_1^2+x_2^2+x_3^2+x_4^2 = \\ = (x_1+x_2+x_3+x_4) ^2 - 2(x_1x_3+x_1x_4+x_2x_3+x_2x_4+x_3x_4)\\ \\ \\ \Rightarrow x_1^4+x_2^4+x_3^4+x_4^4 = 2\cdot (S_1^2-2S_2) - 4 = \\ \\ = 2\cdot \Big(\dfrac{-b}{a}\Big)^2 - 4\cdot \Big(\dfrac{c}{a}\Big)-4 = \\ \\ = 2\cdot \Big(\dfrac{-0}{1}\Big)^2-4\cdot \Big(\dfrac{-2}{1}\Big)-4 = \\ \\ = 2\cdot 0+8-4 = \\ \\ = 4$