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f.4 a maths 一問
If the equations x^2+ac+b=0 and x^2+px+q=0 have a common root, prove that (a-p)(bp-aq)=(b-q)^2. 請詳盡地解釋下.. thx各位好心人^^
最佳解答:
x2+ax+b=0 and x2+px+q=0 have a common root x2+ax+b=0 x2 = ﹣ax﹣b sub into x2+px+q=0 ﹣ax﹣b+px+q=0 (p﹣a)x=b-q x= (b﹣q)/(p﹣a) sub into x2=-ax-b [(b﹣q)/(p﹣a)]2= -a[(b﹣q)/(p﹣a)]﹣b (b﹣q)2= -a(b﹣q)(p-a)-b(p﹣a)2 (b﹣q)2= (p﹣a)[﹣a(b﹣q)﹣b(p﹣a)] (b﹣q)2= (p﹣a)(﹣bp+aq) (a﹣p)(bp﹣aq) = (b﹣q)2
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f.4 a maths 一問
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發問:If the equations x^2+ac+b=0 and x^2+px+q=0 have a common root, prove that (a-p)(bp-aq)=(b-q)^2. 請詳盡地解釋下.. thx各位好心人^^
最佳解答:
x2+ax+b=0 and x2+px+q=0 have a common root x2+ax+b=0 x2 = ﹣ax﹣b sub into x2+px+q=0 ﹣ax﹣b+px+q=0 (p﹣a)x=b-q x= (b﹣q)/(p﹣a) sub into x2=-ax-b [(b﹣q)/(p﹣a)]2= -a[(b﹣q)/(p﹣a)]﹣b (b﹣q)2= -a(b﹣q)(p-a)-b(p﹣a)2 (b﹣q)2= (p﹣a)[﹣a(b﹣q)﹣b(p﹣a)] (b﹣q)2= (p﹣a)(﹣bp+aq) (a﹣p)(bp﹣aq) = (b﹣q)2
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