Question

急求： 已知α、β是方程4x2 -4mx+m+2=0的两个根。求α2 + β2 的最小值

要求过程！谢谢各位大大了

Answer 1

4x^2-4mx+m+2=0
Δ=16m^2-16(m+2)≥0
m^2-m-2≥0

α+β=m αβ=(m+2)/4

α^2+β^2=(α+β)^2-4αβ
=m^2-m-2
=(m-1/2)^2-9/4

α^2+β^2最小值0

Answer 2

α、β是方程4x2 -4mx+m+2=0的两个根
α+β=m
αβ=(m+2)/4

α2+β2=(α+β)2-2αβ=m2-(m+2)/2=m2-m/2+1=m2-m/2+1/16+1-1/16=(m-1/4)2+15/16>=15/16
所以α2 + β2 的最小值是 15/16

Answer 3

方程有实根，判别式≥0
(-4m)²-16(m+2)≥0
m²-m-2≥0
(m-2)(m+1)≥0
m≥2或m≤-1

由韦达定理得
α+β=-(-4m)/4=m
α·β=(m+2)/4
α²+β²=(α+β)²-2α·β=m²-(m+2)/2=m²-m/2 -1=(m-1/4)²-17/16
当m=-1时，α²+β²有最小值1/2

Answer 4

α2 + β2 =(α+β）²-2αβ
=[-（-4m）/4]²-2（m+2）/4
=m²-m/2-1
=（m-1/4）²-17/16
而△≥0
即（-4m）²-4×4×（m+2）≥0
16m²-16m-32≥0
m²-m-2≥0
m1≤-1. m2≥2
∴求α2 + β2 的最小值
α2 + β2=
=（m-1/4）²-17/16
=（-1-1/4）²-17/16
=1/2

Answer 5

方程有根:
判别式≥0

判别式=(-4m)²-4*4*(m+2)=16(m+1)(m-2)≥0
m≥2 或 m≤-1

韦达定理,两根之和为:
α+β=-(-4m)/4=m

求(α2 + β2) ≤-1 或 (α2 + β2)≥2

Answer 6

α2 + β2 = (α + β)^2 - 2αβ = m^2 - 2*(m+2)/4 =
m^2 - (1/2)m + 1 = m^2 - (1/2)m + (1/4)^2 + 15/16 =
(m - 1/4)^2 + 15/16 >= 15/16

有根，需要：
16m^2 - 16(m+2) >= 0
m^2 - m -2 >= 0
m >= 2 , or, m <= -1

于是，
(m - 1/4)^2 + 15/16 >= (-1 - 1/4)^2 + 15/16 = 2.5

α2 + β2 的最小值 = 2.5