A.maths 急急 10分

let z be the common root

z^2+az+b=0.....(1)
z^2+pz+q=0.....(2)

(1) - (2)
(a-p)z + (b-q) = 0
z = -(b-q)/(a-p)....(3)

sub (3) into (1)
[-(b-q)/(a-p)]^2 + a[-(b-q)/(a-p)] + b=0
(b-q)^2 + a(a-p)[-(b-q)] + b(a-p)^2 =0
(b-q)^2 - a(a-p)(b-q) + b(a-p)^2 = 0
a(a-p)(b-q) - b(a-p)^2 = (b-q)^2
(a-p)[a(b-q) - b(a-p)] = (b-q)^2
(a-p)[ab-aq-ab+bp] = (b-q)^2
(a-p)(bp-aq) = (b-q)^2

let c,d be roots ofx^2+ax+b=0
let c,e be roots ofx^2+px+q=0

c+d=-a
a=-c-d

cd=b

c+e=-p
p=-c-e

ce=q

LHS=(a-p)(bp-aq)=(-c-d+c+e)(cd(-c-e)-(-c-d)ce)
=(e-d)(-c^2d-cde+c^2e+cde)
=(e-d)(c^2e-c^2d)=c^2(e-d)^2

RHS=(b-q)^2=(cd-ce)^2
=(ce-cd)^2=c^2(e-d)^2

since LHS=RHS,so(a-p)(bp-aq)=(b-q)^2

Let the common root be m
Therefore, m^2 + am + b = 0 --- (1)
and m^2 + pm + q = 0 --- (2)
(1) - (2)
(a-p)m + b-q = 0
m = (q-b)/(a-p)
Put back value of m to (1)
[(q-b)/(a-p)]^2 + a(q-b)/(a-p) + b = 0
(q-b)^2 + a(q-b)(a-p) + b(a-p)^2 = 0
(q-b)^2 = (b-q)^2 = a(b-q)(a-p) - b(a-p)^2
= (a-p)[a(b-q)-b(a-p)]
= (a-p)(ab-aq-ab+bp)
= (a-p)(bp-aq) (Q.E.D.)

設兩條方程的common root為k,則
k^2 + ak + b = k^2 +pk + q = 0

ak + b = pk + q
(a - p)k = q - b

所以 k = (q-b)/(a-p)

即 (q-b)^2/(a-p)^2 + a(q-b)/(a-p) + b = 0
所以 (q-b)^2 + a(q-b)(a-p) + b(a-p)^2 = 0
(b-q)^2 = [a(b-q) - b(a-p)](a-p)
= (ab - aq - ba + bp)(a-p)
= (bp - aq)(a - p)

2006-11-01 22:25:50 補充：
頭兩行算式有亂碼:k^2 ak b = k^2 kp q = 0ak b = kp q