F.4 a math 問題(2)

Let c be the common root,
c^2 + ac + b = 0 --------------- Equation1
c^2 + pc + q = 0 --------------- Equation2
Equation1 - Equation2 = (a-p)c= (q-b)
c = (q-b)/(a-p) ------------------ Equation3
Sub Equation3 into Equation1,
(q-b)^2/(a-p)^2 + a(q-b)/(a-p) + b = 0
(q-b)^2 + a(q-b)(a-p) + b(a-p)^2 = 0
(a-p)(aq-ab+ba-bp) + (b-q)^2 = 0
(a-p)(bp-aq) = (b-q)^2

Let z be the common root of two equations ........

therefore,we have:
z^2+az+b=0...................1
z^2+pz+q=0...................2

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

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