標題:
A.maths 急急 10分
發問:
有兩條野.. x^2+ax+b=0 x^2+px+q=0 佢地都有一個common root Prove (a-p)(bp-aq)=(b-q)^2 我要Steps +答案.... THZ@@
最佳解答:
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
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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
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