F.4 maths 2題 20分!!急

5.
assume α,β are the roots of the equation: ax^2+bx+c=0, so
α+β=-b/a.................i
α*β=c/a...................ii
Also, α,β are the reciprocals of the roots of the equation: dx^2+ex+f=0, so
(1/α)+(1/β)=-e/d.................iii
(1/α)*(1/β)=f/d...................iv
from ii and iv, you should be able to find ad=cf..............v
from i, iii and v, you should be able to find ae=bf.

6.
α,β are the roots of the equation: x^2-x-1=0, so
α+β=1 =>α=1-β
α*β=-1 => (1-β)*β=-1 => β^2=1+β...................i
vice versa, you should be able to find α^2=1+α.................ii
f(n)+f(n+1)
=1/√5(α^n-β^n)+1/√5[α^(n+1)-β^(n+1)]
simplify it, then you should be able to get:
=1/√5[(α^n)(1+α)-(β^n)(1+β)]......................................iii
substitute i & ii to iii,
you should be able to get:
=1/√5[(α^n)(α^2)-(β^n)(β^2)]
=1/√5[α^(n+2)-β^(n+2)]
=f(n+2)

Have fun to follow the steps and finish the gap. You will be proud of yourself. :-)

Notes:
The key formulas for those two questions are:
if α,β are the roots of the equation: ax^2+bx+c=0
α+β=-b/a
α*β=c/a
if α,β the reciprocal of the roots (s,t) of dx^2+ex+f=0
α=1/s
β=1/t
therefore,
(1/α)+(1/β)=-e/d
(1/α)*(1/β)=f/d

5. Let α, β be the roots of the equation dx²＋ex＋f＝0, so α＋β＝-e/d and αβ＝f/d.
As the roots of the equation ax²＋bx＋c＝0 are 1/α and 1/β, so,
1/α＋1/β＝-b/a
==> (α+β)/αβ＝-b/a
==> (-e/d)/(f/d)＝-b/a
==> ae＝bf
(1/α)(1/β)＝c/a
==> 1/αβ＝c/a
==> 1/(f/d)＝c/a
==> ad＝cf
∴ ae＝bf, ad＝cf.

6. As α, β are the roots of the equation x²－x－1＝0, so, α＋β＝1 and αβ＝-1.
As α＞β, so,
α－β＝√[(α＋β)²－4αβ]＝√5
f(n+2)
＝(1/√5) [α^(n+2)－β^(n+2)]
＝(1/√5) {[α^(n+1)－β^(n+1)](α＋β)－βα^(n+1)＋αβ^(n+1}
＝(1/√5) [α^(n+1)－β^(n+1)]－(1/√5) αβ(α^n－β^n)
＝(1/√5) [α^(n+1)－β^(n+1)]＋(1/√5) (α^n－β^n)
＝f(n+1)＋f(n)