均是利用方程ax²+bx+c=0两根之和等于-b/a,两根之积等于c/a来解题
1.x1+x2=2,则x1+3x2=2+2x2=3,解出x2=0.5,代入原方程,解出m=0.75
2.x1+x2=1,x1x2=p-1
则[2+x1(1-x1][2+x2(1-x2)]=9=4+2(x1+x2-(x1+x2)²+2x1x2)+x1x2[1+x1x2-(x1+x2)]=4p+(p-1)²=(p+1)²
所以p=2或-4,又要考虑Δ>0,则舍去2,取p=-4
均是利用方程ax²+bx+c=0两根之和等于-b/a,两根之积等于c/a来解题
1.x1+x2=2,则x1+3x2=2+2x2=3,解出x2=0.5,代入原方程,解出m=0.75
2.x1+x2=1,x1x2=p-1
则[2+x1(1-x1][2+x2(1-x2)]=9=4+2(x1+x2-(x1+x2)²+2x1x2)+x1x2[1+x1x2-(x1+x2)]=4p+(p-1)²=(p+1)²
所以p=2或-4,又要考虑Δ>0,则舍去2,取p=-4