\(\Delta = {b^2} - 4ac = {\left( { - 4} \right)^2} - 4.1.\left( {m + 3} \right) = 4 - 4m\)
Để phương trình có hai nghiệm phân biệt:
\(\Delta > 0 \Rightarrow 4 - 4m > 0 \Rightarrow m < 1\)
Theo hệ thức Vi - ét: \(\left\{ \begin{array}{l} {x_1} + {x_2} = - \dfrac{b}{a}\\ {x_1}{x_2} = \dfrac{c}{a} \end{array} \right. \Leftrightarrow \left\{ \begin{array}{l} {x_1} + {x_2} = 4\\ {x_1}{x_2} = m + 3 \end{array} \right.\)
Ta có:
\(\begin{array}{l} \left| {{x_1} - {x_2}} \right| = 2\\ \Leftrightarrow \left( {\left| {{x_1} - {x_2}} \right|} \right) = {2^2}\\ \Leftrightarrow {\left( {{x_1} + {x_2}} \right)^2} - 4{x_1}{x_2} = 4\\ \Leftrightarrow {4^2} - 4.\left( {m + 3} \right) = 4\\ \Leftrightarrow 16 - 4m - 12 = 4\\ \Leftrightarrow m = 0 \text{(thỏa mãn)} \end{array}\)
Vậy \(m=0\)
\(\Delta'=4-\left(m+3\right)=1-m>0\Rightarrow m< 1\)
Theo định lý Viet: \(\left\{{}\begin{matrix}x_1+x_2=4\\x_1x_2=m+3\end{matrix}\right.\)
\(\left|x_1-x_2\right|=2\)
\(\Leftrightarrow\left(x_1-x_2\right)^2=4\)
\(\Leftrightarrow\left(x_1+x_2\right)^2-4x_1x_2=4\)
\(\Leftrightarrow16-4\left(m+3\right)=4\)
\(\Rightarrow m=0\) (thỏa mãn)