Question

Cho pt bậc hai : x2 - 2mx + 4m -4 = 0 (1)
Tìm m để 2 nghiệm x₁,x₂ thỏa hệ thức (1) : (x₁- 2)(x₂ - 2)= x₁² + x₂² - 8

Answer 1

\(\Delta'=m^2-4m+4=\left(m-2\right)^2\ge0\)

Phương trình luôn có nghiệm, theo Viet: \(\left\{{}\begin{matrix}x_1+x_2=2m\\x_1x_2=4m-4\end{matrix}\right.\)

\(\left(x_1-2\right)\left(x_2-2\right)=x_1^2+x_2^2-8\)

\(\Leftrightarrow x_1x_2-2\left(x_1+x_2\right)+4=\left(x_1+x_2\right)^2-2x_1x_2-8\)

\(\Leftrightarrow\left(x_1+x_2\right)^2-3x_1x_2+2\left(x_1+x_2\right)-12=0\)

\(\Leftrightarrow4m^2-3\left(4m-4\right)+4m-12=0\)

\(\Leftrightarrow4m^2-8m=0\Rightarrow\left[{}\begin{matrix}m=0\\m=2\end{matrix}\right.\)