Question

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

Tìm m để phương trình có 2 nghiệm x1, x2 thỏa mãn x1 = 2x₂

Answer 1

Theo he thuc Viet ta co:

\(x_1+x_2=-\frac{b}{a}=\frac{2\left(m+1\right)}{1}=2\left(m+1\right)\)

\(x_1\cdot x_2=\frac{c}{a}=\frac{m^2+1}{1}=m^2+1\)

Theo y/c de bai \(x_1=2x_2\) thay x1 vao hai he thuc tren ta co:

\(\left\{{}\begin{matrix}3x_2=2\left(m+1\right)\\2x_2 ^2=m^2+1\end{matrix}\right.\)<=>\(\left\{{}\begin{matrix}9x^2_2=4\left(m+1\right)^2\\2x^2_2=m^2+1\end{matrix}\right.\Rightarrow\frac{9}{2}=\frac{4\left(m+1\right)^2}{m^2+1}\)

\(\Leftrightarrow9m^2+9=8\left(m^2+2m+1\right)\Leftrightarrow m^2-16m+1=0\Rightarrow\left[{}\begin{matrix}m=8+3\sqrt{7}\\m=8-3\sqrt{7}\end{matrix}\right.\)

Vay co 2 gia tri m thoa man ycbt.