Đáp án:
Theo vi-ét:
$\left \{ {{x_1.x_2=\frac{c}{a}} \atop {x_1+x_2=\frac{-b}{a}}} \right.$
⇔$\left \{ {{x_1.x_2=m+1} \atop {x_1+x_2=4}} \right.$
Ta có:
$x_1{^2}$+$x_2{^2}$=$5$($x_1$+$x_2$)
⇔$x_1{^2}$+$2$$x_1$$x_2$+$x_2{^2}$-$2$$x_1$$x_2$=$5$($x_1$+$x_2$)
⇔$(x_1+x_2)^2$-$2$$x_1$$x_2$=$5$($x_1$+$x_2$)
⇔$(4)^2$ -$2.(m+1)$=$5.4$
⇔$16$$-2(m+1)$=$20$
⇔$-2(m+1)$=$4$
⇔$m+1$=$-2$
⇔$m=-3$
