Question

Tìm m để phương trình \(x^2-4x+m+2=0\) có 2 nghiệm x₁; x₂ thỏa mãn \(\left(x_1\right)^3+\left(x_2\right)^3=28\)

Answer 1

\(x^2-4x+m+2=0\)

Theo định lý Viet

\(\Rightarrow\left\{{}\begin{matrix}x_1+x_2=\dfrac{-b}{a}\\x_1x_2=\dfrac{c}{a}\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x_1+x_2=4\\x_1x_2=m+2\end{matrix}\right.\)

Theo đề bài ta có \(x_1^3+x_2^3=28\)

\(\Leftrightarrow\left(x_1+x_2\right)^3-3x_1x_2\left(x_1+x_2\right)=28\)

\(\Leftrightarrow4^3-12\left(m+2\right)=28\)

\(\Leftrightarrow12\left(m+2\right)=36\)

\(\Leftrightarrow m+2=3\)

\(\Leftrightarrow m=1\)