Question

Cho PT: \(x^2-2\left(m-1\right)x-m-3=0\)(1)

Tìm m để PT (1) có 2 nghiệm thỏa mãn hệ thức \(x_1^2+x_2^2=10\)

Answer 1

\(\Delta=\left[-2\left(m-1\right)\right]^2-4\left(-m-3\right)\)

   \(=4\left(m^2-2m+1\right)+4m+12=4m^2-8m+4+4m+12=4m^2-4m+16=\left(2m-1\right)^2+15>0;\forall m\)

=> pt luôn có 2 nghiệm phân biệt

Theo hệ thức Vi-ét, ta có: \(\left\{{}\begin{matrix}x_1+x_2=2m-2\\x_1.x_2=-m-3\end{matrix}\right.\)

\(x_1^2+x_2^2=10\)

\(\Leftrightarrow\left(x_1+x_2\right)^2-2x_1.x_2=10\)

\(\Leftrightarrow\left(2m-2\right)^2-2\left(-m-3\right)=10\)

\(\Leftrightarrow4m^2-8m+4+2m+6=10\)

\(\Leftrightarrow4m^2-6m=0\)

\(\Leftrightarrow2m\left(2m-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}m=0\\m=\dfrac{3}{2}\end{matrix}\right.\)