Question

[Na Na]

Tìm m để 2 pt sau có nghiệm chung:

x^2 - 2mx - 4m + 1 = 0 (1)

x^2 + (3m+1)x + 2m + 1 = 0 (2)

Answer 1

[Trần Tuấn Hoàng]

\(x^2-2mx-4m+1=0\left(1\right)\)

\(x^2+\left(3m+1\right)x+2m+1=0\left(2\right)\)

Gọi x₀ là nghiệm chung của hai phương trình trên. Do đó ta có:

\(\left\{{}\begin{matrix}x_0^2-2mx_0-4m+1=0\left(3\right)\\x_0^2+\left(3m+1\right)x_0+2m+1=0\end{matrix}\right.\)

\(\Rightarrow\left(3m+1\right)x_0+2m+1-\left(-2mx_0-4m+1\right)=0\)

\(\Rightarrow\left(5m+1\right)x_0+6m=0\)

\(\Rightarrow m\left(5x_0+6\right)+x_0=0\)

\(\Rightarrow m=\dfrac{-x_0}{5x_0+6}\) \(\left(x_0\ne\dfrac{-6}{5}\right)\)

Thay vào (3) ta được:

\(x_0^2-2.\dfrac{-x_0}{5x_0+6}.x_0-4.\dfrac{-x_0}{5x_0+6}+1=0\)

\(\Rightarrow x_0^2+\dfrac{2x_0^2}{5x_0+6}+\dfrac{4x_0}{5x_0+6}+1=0\)

\(\Leftrightarrow x_0^2\left(5x_0+6\right)+2x_0^2+4x_0+5x_0+6=0\)

\(\Leftrightarrow5x_0^3+8x_0^2+9x_0+6=0\)

\(\Leftrightarrow5x_0^3+5x_0^2+3x_0^2+3x_0+6x_0+6=0\)

\(\Leftrightarrow5x_0^2\left(x_0+1\right)+3x_0\left(x_0+1\right)+6\left(x_0+1\right)=0\)

\(\Leftrightarrow\left(x_0+1\right)\left(5x_0^2+3x_0+6\right)=0\)

\(\Leftrightarrow x_0=-1\)

\(\Rightarrow m=\dfrac{-x_0}{5x_0+6}=\dfrac{-\left(-1\right)}{5.\left(-1\right)+6}=\dfrac{1}{6}\)

Answer 2

[Nguyễn Huy Tú]

Xét (1) : Để pt có nghiệm khi 

\(\Delta'=m^2-\left(-4m+1\right)=m^2+4m-1\ge0\)

\(\Leftrightarrow\left[{}\begin{matrix}x\le-2-\sqrt{5}\\x\ge-2+\sqrt{5}\end{matrix}\right.\)

(2) : Để pt có nghiệm khi \(\Delta=\left(3m+1\right)^2-4\left(2m+1\right)=9m^2+6m+1-8m-4=9m^2-2m-3\ge0\Leftrightarrow\left[{}\begin{matrix}x\le\dfrac{1-2\sqrt{7}}{9}\\x\ge\dfrac{1+2\sqrt{7}}{9}\end{matrix}\right.\)

Để 2 pt có nghiệm chung khi \(\left[{}\begin{matrix}x\le-2-\sqrt{5}\\x\ge\dfrac{1+2\sqrt{7}}{9}\end{matrix}\right.\)