標題:
F.4 a math 問題(2)
發問:
if the equations x^2 + ax + b =0 and x^2 +px+q =0 have a common root, prove that (a - p)(bp - aq)=(b - q)^2
最佳解答:
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
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其他解答:
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)
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