试题:
当k=______时,关于x的一元二次方程kx2-2(k+1)x+k-1=0的两个不相等的实数根x1,x2满足
1
x1
+
1
x2
=3

答案:

我来补答
根据题意得
x1+x2=-
b
a
=-
-2(k+1)
k
=
2(k+1)
k

x1x2=
c
a
=
k-1
k

又∵
1
x1
+
1
x2
=3,
1
x1
+
1
x2
=
x1+x2
x1x2
=
2(k+1)
k
×
k
k-1
=
2(k+1)
k-1
=3,
即2(k+1)=3(k-1),
解得k=5.
 
 
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