Question

Cho phương trình : \(x^2-2\left(m+1\right)x-12=0\)

Tìm m để phương trình có nghiệm x1, x2 thỏa mãn : \(x_1^2-x_2^2-7\cdot2\cdot\left(m+1\right)=0\)

Answer 1

\(ac=-12< 0\) nên pt luôn có 2 nghiệm pb trái dấu

Theo hệ thức Viet: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m+1\right)\\x_1x_2=-12\end{matrix}\right.\)

\(x_1^2-x_2^2-14\left(m+1\right)=0\)

\(\Leftrightarrow\left(x_1-x_2\right)\left(x_1+x_2\right)-14\left(m+1\right)=0\)

\(\Leftrightarrow\left(x_1-x_2\right).2\left(m+1\right)-14\left(m+1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}m=-1\\x_1-x_2=7\left(1\right)\end{matrix}\right.\)

Xét (1), kết hợp với Viet ta được: \(\left\{{}\begin{matrix}x_1+x_2=2\left(m+1\right)\\x_1-x_2=7\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}x_1=\dfrac{2m+9}{2}\\x_2=\dfrac{2m-5}{2}\end{matrix}\right.\)

Thế vào \(x_1x_2=-12\Leftrightarrow\left(\dfrac{2m+9}{2}\right)\left(\dfrac{2m-5}{2}\right)=-12\)

\(\Leftrightarrow4m^2+8m+3=0\Rightarrow\left[{}\begin{matrix}m=-\dfrac{3}{2}\\m=-\dfrac{1}{2}\end{matrix}\right.\)

Vậy \(m=\left\{-1;-\dfrac{3}{2};-\dfrac{1}{2}\right\}\)