Question

X2+（2k-1）x+k2=0的两个根x1的平方-x2的平方=0

Answer 1

(x1)^2-(x2)^2=0

(x1+x2)(x1-x2)=0

-(2k-1)(x1-x2)=0

(2k-1)(x1-x2)=0

2k-1=0或x1-x2=0

k=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

-(4k-1)=0

4k-1=0

k=1/4