Question

已知方程x^2+(2k+1)x+k-1=0的两个实数根分别为x1 x2

且满足x1-x2=4k-1,求实数k的值

Answer 1

x1+x2=-(2k+1)
x1x2=k-1

x1-x2=4k-1
(x1-x2)^2=(4k-1)^2
(x1+x2)^2-4x1x2=16k^2-8k+1
4k^2+4k+1-4k+4=16k^2-8k+1
12k^2-8k-4=0
3k^2-2k-1=0
(3k+1)(k-1)=0
k=-1/3 or 1

Answer 2

x₁+x₂=-(2k+1)

x₁-x₂=4k-1
∴x₁=k-1
x₂=-3k;
∵x₁x₂=k-1
∴(k-1)(-3k)=k-1
∴(k-1)(-3k-1)=0
∴k=1或k=-1/3