x21+x22−3(x1+x2)=16x12+x22−3(x1+x2)=16
⇔(x1+x2)2−2x1x2−3(x1+x2)=16⇔(x1+x2)2−2x1x2−3(x1+x2)=16
Theo định lí Vi-ét:
x1+x2=m−1x1+x2=m−1
x1x2=−2m−3x1x2=−2m−3
⇒(x1+x2)2−2x1x2−3(x1+x2)=16⇒(x1+x2)2−2x1x2−3(x1+x2)=16
⇔(m−1)2−2(−2m−3)−3(m−1)−16=0⇔(m−1)2−2(−2m−3)−3(m−1)−16=0
⇔m2−2m+1+4m+6−3m+3−16=0⇔m2−2m+1+4m+6−3m+3−16=0
⇔m2−m−6=0⇔m2−m−6=0
⇔[m=3x=−2⇔[m=3x=−2.
Vậy m=3m=3 hoặc m=−2m=−2.