Question

已知x1,x2是关于x的一元二次方程x^2 - 2(m+2)x + 2m^2 - 1=0的两个实数根，且满足x1^2 - x2^2=0，求m的值

Answer 1

根据韦达定理有：

X1 + X2 = 2(m+2)

而x1^2 - x2^2=0

则x1^2 - x2^2 = (X1 + X2 )(X1 - X2) = 0

X1 + X2 = 0
X1 - X2 = 0

由X1 + X2 = 0得:2(m+2)=0..................m = -2
由X1 - X2 = 0得:判别式=0...................4(m+2)^2 - 4(2m^2 - 1)=0..........m = 5或 -1

当m = -2时,原方程:x^2 + 7=0无实数根,所以舍去.

综上,m = 5或 -1.

Answer 2

由题，x1^2 - x2^2=0
即 x1+x2=0 或x1-x2=0
x1+x2=2(m+2)=0 ==> m=-2
x1-x2=√(4*(m+2)^2-4*(2m^2-1))=√-4m^2+16m+20=0
解得m=5 或m=-1