已知方程x^2-2ax+(a^2-a+6)=0的两个根为α,β,求（α-1）^2+(β-1)^2的最小值

Question

已知方程x^2-2ax+(a^2-a+6)=0的两个根为α,β,求（α-1）^2+(β-1)^2的最小值

Answer 1

（α-1）^2+(β-1)^2
=a²-2a+1+b²-2b+1
=a²+b²-2(a+b)+1
=(a+b)²-2ab-2(a+b)+1
=(2a)²-2(a²-a+6)-2*2a+1
=4a²-2a²+2a-12-4a+1
=2a²-2a-11
=2(a²-a)-11
=2(a-0.5)²-11.5
当a=0.5时,有是小值为-11.5