Question

已知x1,x2是方程x^2-2(k+1)x+K^2+2=0的两个实数根，且(x1+1)(x2+1)=8,求K的值

Answer 1

根据“根与系数的关系”，可得x1+x2=2(k+1),x1x2=k^2+2
(x1+1)(x2+1)
=x1x2+x1+x2+1
=k^2+2+2(k+1)+1
=k^2+2k+5
∴k^2+2k+5=8,解得k=1或-3
判别式△=4(k+1)^2-4(k^2+2)=4k^2+8k+4-4k^2-8=8k-4>0
∴k=-3舍去，k=1

Answer 2

x1+x2=2k+2
x1x2=K^2+2
(x1+1)(x2+1)
=x1x2+x1+x2+1
=K^2+2+2k+2+1
=K^2+2k+5=8
K^2+2k-3=0
(k+3)(k-1)=0
k=1 k=-3
b^2-4ac
=4(k+1)^2-4(K^2+2)
=4k^2+8k+4-4k^2-8
=8k-4>0
k>1/2
∴k=1

Answer 3

x1x2=k²+2
x1+x2=2(k+1)

(x1+1)(x2+1)=8
x1x2+(x1+x2)+1=8
k²+2+2(k+1)+1-8=0
k²+2k-3=0
k=-3 （舍去） k=1