Question

已知α,β是方程x^2 -x-1=0的两个实数解求2α^2+5β^2

韦达定理

Answer 1

α+β=1
αβ=-1
(α-β)²=(α+β)²-4αβ=5，所以β-α=±√5
α²=α+1，β²=β+1
2α²+5β²=2α+5β+7=3β+9
2α²+5β²=2α+5β+7=-3α+12.

2[2α²+5β²]=3(β-α)+21=21±3√5.
所以2α²+5β²=(21±3√5)/2.

Answer 2

x^2 -x-1=0
x = (1 ± √5)/2

case 1: α = (1 + √5)/2 , β=(1 -√5)/2
2α^2+5β^2
=2(1 + √5)/2)^2+5(1 -√5)/2)^2
= (1/4)[ 2(6+2√5) +5(6-2√5)]
=(1/2)[ 21 -3√5]

case 2: α = (1 - √5)/2 , β=(1 +√5)/2
2α^2+5β^2
=2(1 - √5)/2)^2+5(1+√5)/2)^2
= (1/4)[ 2(6-2√5) +5(6+2√5)]
=(1/2)[ 21+3√5]

ie
2α^2+5β^2=(1/2)[ 21 ± 3√5]