6.若a>0,b>0,α、β為x^2-ax+b=0的兩根,且α+β=α^2+β^2=α^3+β^3,求a、b的值。
39640366 發表於 2011-12-11 13:49 
α、β為x²-ax+b = 0的兩根
∴α+β = a , αβ = b
α+β = α²+β² = α³+β³ = a
α²+β² = (α+β)²-2αβ = a²-2b
α³+β³ = (α+β)³-3α²β-3αβ² = (α+β)³-3αβ(α+β) = a³-3ab
∴ a = a²-2b = a³-3ab
2b = a²-a
2a = 2a³-3a(2b) = 2a³-3a(a²-a) ,a>0
2 = 2a²-3(a²-a) = -a²+3a
a²-3a+2 = 0 a = 1 or 2
a = 1時
2b = a²-a = 0 → b = 0 (b>0不合)
a = 2時
2b = a²-a = 2 → b = 1
Ans: a=2, b=1
驗証:
a = 2 , a²-2b = 2 , a³-3ab = 2
α、β為x²-2x+1 = 0的兩根
α = 1 , β = 1
1+1 = 1²+1² = 1³+1³ = 2 |