Question

помогите алгебра номер 1​

Answer 1

Ответ:     $x_1^2+x_2^2+x_3^2=32\ .$

$x^3+px^2+qx+r=0\ \ \Rightarrow \ \ \left\{\begin{array}{l}x_1+x_2+x_3=-p\\x_1x_2+x_1x_3+x_2x_3=q\\x_1x_2x_3=-r\end{array}\right\ \ (teorema\ Vieta)\\\\\\x^3+2x^2-14x-3=0\ \ \Rightarrow \ \ \left\{\begin{array}{l}x_1+x_2+x_3=-2\\x_1x_2+x_1x_3+x_2x_3=-14\\x_1x_2x_3=3\end{array}\right$

$(x_1+x_2+x_3)^2=(\, (x_1+x_2)+x_3\, )^2=(x_1+x_2)^2+x_3^2=\\\\=(x_1^2+2x_1x_2+x_2^2)+2(x_1+x_2)\, x_3+x_3^2=\\\\=x_1^2+x_2^2+x_3^2+2\, (x_1x_2+x_1x_3+x_2x_3)\ \ \ \Rightarrow \\\\\boxed {\ x_1^2+x_2^2+x_3^2=(x_1+x_2+x_3)^2-2\, (x_1x_2+x_1x_3+x_2x_3)\ }\\\\\\x_1^2+x_2^2+x_3^2=(-2)^2-2\cdot (-14)=4+28=32$