实数a,b满足a²-2a-5=0,5b²+2b-1=0,求a²+1/b²

实数a,b满足a²-2a-5=0,5b²+2b-1=0,求a²+1/b²
实数a,b满足a²-2a-5=0,5b²+2b-1=0,求a²+1/b²

实数a,b满足a²-2a-5=0,5b²+2b-1=0,求a²+1/b²
a²-2a-5=0,
5b²+2b-1=0,
5+2/b-1/b^2=0
1/b^2-2/b-5=0
所以a,1/b是关于方程x^2-2x-5=0的两个根
a+1/b=2
a*1/b=-5
a²+1/b²
=a²+1/b²+2*a*1/b-2*a*1/b
=(a+1/b)²-2*(-5)
=2²+10
=14