Question

已知关于x的方程x的平方+(2k+1)x+k的平方-2=0的两实根的平方和等于11,则k的值为?

Answer 1

嘿嘿
假设2个根分别为x1,x2 则
x1+x2=-(2k+1)
x1x2=k²-2
x1²+x2²=(x1+x2)²-2x1x2=(2k+1)²-2(k²-2)=2k²+4k+5=11
2k²+4k-6=0
(k+3)(2k-2)=0
k=-3 or k=1
△=(2k+1)²-4(k²-2)=4k+9≥0
k≥-9/4
k=-3舍去
k=1

Answer 2

x1+x2=-(2k+1)
x1*x2=k^2-2
x1^2+x2^2=(x1+x2)^2-2x1x2=(2k+1)^2-2*(k^2-2)=2k^2+4k+5=11
2k^2+4k-6=0
k^2+2k-3=0
(k+3)(k-1)=0
k=-3,k=1
（k+1)^2-4(k^2-2)
=4k^2+4k+1-4k^2+8
=4k+9≥0
所以k≥-9/4所以k=1

Answer 3

x²+(2k+1)x+k²-2=0
⊿=(2k+1)²-4(k²-2)＞0
4k＞-9
k＞-9/4
x1+x2=-(2k+1)
x1·x2=k²-2
x1²+x2²=(x1+x2)²-2x1·x2
11=(2k+1)²-2(k²-2)
k²+2k-3=0
(k+3)(k-1)=0
k1=-3, k2=1
∵k＞-9/4
∴取k=1

Answer 4

x1+x2=-(2k+1)
x1*x2=k^2-2
x1^2+x2^2=(x1+x2)^2-2x1x2=(2k+1)^2-2*(k^2-2)=2k^2+4k+5=11
2k^2+4k-6=0
k^2+2k-3=0
(k+3)(k-1)=0
k=-3,k=1
△=（2k+1)^2-4(k^2-2)
=4k^2+4k+1-4k^2+8