Question

已知a,b是方程4x^2-4mx+m+2=0的两个实数根,求(a-1)^2+(b-1)^2的最小值.急

Answer 1

方程有实数根,则△≥0

16m²-16(m+2)≥0

m²-m-2≥0

(m+1)(m-2)≥0

m≥2或m≤-1

根据韦达定理,

a+b=4m/4=m

a*b=(m+2)/4

(a-1)²+(b-1)²

=(a²+b²)-2(a+b)+2

=(a+b)²-2ab-2(a+b)+2

=m²-(m+2)/2-2m+2

=m²-2.5m+1

=(m-1.25)²+1-1.5625

=(m-1.25)²-0.5625

当m=2时有最小值,为:

(2-1.25)²-0.5625=0