Question

cho phương trình \(x^2+4x+1=0\)

tính \(C=\left(x_1\right)^2-\left(x_2\right)^2\)

Answer 1

\(\left\{{}\begin{matrix}x_1+x_2=-4\\x_1x_2=1\end{matrix}\right.\)

\(\left(x_1-x_2\right)^2=\left(x_1+x_2\right)^2-4x_1x_2=\left(-4\right)^2-4\cdot1=12\)

\(\Leftrightarrow\left[{}\begin{matrix}x_1-x_2=2\sqrt{3}\\x_1-x_2=-2\sqrt{3}\end{matrix}\right.\)

\(C=\left(x_1+x_2\right)\left(x_1-x_2\right)\)

\(=\left[{}\begin{matrix}-4\cdot2\sqrt{3}=-8\sqrt{3}\\8\sqrt{3}\end{matrix}\right.\)