Question

已知方程x2-2mx+4m2-6=0﹙m∈R﹚的两个实根为αβ求﹙α-1﹚2+﹙β-1﹚2的最大值和最小值

已知方程x²-2mx+4m²-6=0﹙m∈R﹚的两个实根为αβ求﹙α-1﹚²+﹙β-1﹚²的最大值和最小值

x²-2mx+4m²-6=0
Δ=m²-4m²+6≥0
=(m-2)²+2≥0恒成立
α+β=2m , αβ=4m²-6
原式=α²-2α+1+β²-2β+1
=α²+β²+2αβ-2αβ-2(α+β)+1
=(α+β)²-2(αβ+a+b)+1
=4m²-2(4m²-6)+1
= -4m²+25
最大值为

然后呢...

Answer 1

Δ=m²-4m²+6=6-3m²≥0，m²≤2，-√2≤m≤√2
α+β=2m , αβ=4m²-6
原式=α²-2α+1+β²-2β+1
=α²+β²+2αβ-2αβ-2(α+β)+1
=(α+β)²-2(αβ+α+β)+1
=4m²-2(4m²-6+2m)+1
= -4m²-4m+13
-8≤-4m²≤0,-4√2≤-4m≤4√2
-8-4√2≤-4m²-4m≤4√2
所以5-4√2≤-4m²-4m+13≤4√2+13
所以最大值为4√2+13，最小值为5-4√2