a) Để `f(x)` nhận `-2` làm nghiệm
`⇒f(-2)=0`
`⇒(-2)^2+m(-2)+2=0`
`⇔4-2m+2=0`
`⇔6-2m=0`
`⇔2m=6`
`⇔m=3`
b) Với `m=3`
`⇒f(x)=x^2+3x+2`
Đặt `f(x)=0`
`⇒x^2+3x+2=0`
`⇔x^2+2.x. 3/2+9/4-1/4=0`
`⇔(x+3/2)^2=1/4`
`⇔`\(\left[ \begin{array}{l}x+\dfrac{3}{2}=\dfrac{\sqrt{1}}{2}\\x+\dfrac{3}{2}=-\dfrac{\sqrt{1}}{2}\end{array} \right.\)
`⇔`\(\left[ \begin{array}{l}x=\dfrac{\sqrt{1}-3}{2}\\x=\dfrac{-\sqrt{1}-3}{2}\end{array} \right.\)