X2+(2k-1)x+k2=0的两个根x1的平方-x2的平方=0

问题描述:

X2+(2k-1)x+k2=0的两个根x1的平方-x2的平方=0
已知:关于x的一元二次方程x2+(2k-1)x+k2=0的两根x1,x2满足x12-x22=0,双曲线y=
y= y=
4kx(x>0)经过Rt△OAB斜边OB的中点D,与直角边AB交于C(如图),求S△OBC

(x1)^2-(x2)^2=0(x1+x2)(x1-x2)=0-(2k-1)(x1-x2)=0(2k-1)(x1-x2)=02k-1=0或x1-x2=0k=1/2或x1=x2当k=1/2时x^2+(1/2)^2=0不成立 当x1=x2时,x^2+(2k-1)x+k^2=0△=(2k-1)^2-4k^2=0(2k-1)^2-4k^2=0(2k-1-2k)(2k-1+2k)=0-(...