Question

已知关于x的一元二次方程X^2-2X-2KX+K^2=0的两个实数根为X1,X2

Answer 1

X^2-2X-2KX+K^2=0

x^2-(2+2k)x+k^2=0

x1+x2=2+2k,x1x2=k^2

∆=4(k+1)^2-4k^2=8k+4>=0

则k>=-1/2

1.y=x1+x2=2+2k>=1

2.Y=X1+X2+√X1*X2

=2+2k-k

=k+2

y-(1-3k)=k+2-1+3k=4k+1

若y>1-3k,即4k+1>0,则-1/4