\(ax^2+bx+c=0\)
Theo định lý Viet
\(\Rightarrow\left\{{}\begin{matrix}S=x_1+x_2=\dfrac{-b}{a}\\P=x_1x_2=\dfrac{c}{a}\end{matrix}\right.\)
Ta có pt bậc 2 có 2 nghiệm là \(\dfrac{1}{x^2_1};\dfrac{1}{x^2_2}\)
\(\Rightarrow\left\{{}\begin{matrix}S'=\dfrac{1}{x^2_1}+\dfrac{1}{x^2_2}\\P'=\dfrac{1}{x^2_1x^2_2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}S'=\dfrac{\left(x_1+x_2\right)^2-2x_1x_2}{x^2_1x^2_2}\\P'=\dfrac{1}{x^2_1x^2_2}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}S'=\dfrac{\left(\dfrac{-b}{a}\right)^2-\dfrac{2c}{a}}{\left(\dfrac{c}{a}\right)^2}\\P'=\dfrac{1}{\left(\dfrac{c}{a}\right)^2}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}S'=\dfrac{\dfrac{b^2}{a^2}-\dfrac{2c}{a}}{\dfrac{c^2}{a^2}}\\P'=\dfrac{1}{\dfrac{c^2}{a^2}}\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}S'=\dfrac{\dfrac{b^2-2ca}{a^2}}{\dfrac{c^2}{a^2}}=\dfrac{b^2-2ca}{c^2}\\P'=\dfrac{a^2}{c^2}\end{matrix}\right.\)
Theo định lý Viet đảo pt bậc 2 cần lập
\(\Leftrightarrow z^2-S'z+P'=0\)
\(\Leftrightarrow z^2-\dfrac{b^2-2ca}{c^2}z+\dfrac{a^2}{c^2}=0\)
