Question

初中一道方程组的题已知方程组y^2=2x,y=x+m,的两组解分别是x=x1,y=y1;和x=x2,y=y2.且1/x1

Answer 1

y=x+m

y²=x²+2mx+m²=2x

x²+(2m-2)x+m²=0

韦达定理

x1+x2=-(2m-2)

x1x2=m²

1/x1+1/x2

=(x1+x2)/x1x2=3/2

-(2m-2)/m²=3/2

3m²+4m-4=0

(3m-2)(m+2)=0

m=2/3,m=-2

有解的判别式大于等于0

(2m-2)²-4m²>=0

m=2/3不成立

所以m=-2