Question

(x-m+1)(x+2m-3)=0, tìm m để pt có 2 nghiệm phân biệt

Answer 1

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

\(\Leftrightarrow x^2+2xm-3x-xm-2m^2+3m+x+2m-3=0\)

\(\Leftrightarrow-2m^2+m\left(x+5\right)+\left(x^2-2x-3\right)=0\)

Để pt có 2 nghiệm phân biệt thì \(\Delta>0\)

\(\Leftrightarrow\left(x+5\right)^2-4\left(x^2-2x-3\right)\left(-2\right)>0\)

\(\Leftrightarrow x^2+10x+25+8\left(x^2-2x-3\right)>0\)

\(\Leftrightarrow x^2+10x+25+8x^2-16x-24>0\)

\(\Leftrightarrow9x^2-6x+1>0\)

\(\Leftrightarrow\left(3x-1\right)^2>0\)

\(\Leftrightarrow\left|3x-1\right|>0\)

\(\Leftrightarrow\left[{}\begin{matrix}3x-1< 0\\3x-1>0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x< \frac{1}{3}\\x>\frac{1}{3}\end{matrix}\right.\)

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