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^^
f.4 a maths 一問
2007-11-26 1:44 am
回答 (1)
2007-11-26 1:51 am
✔ 最佳答案
x2+ax+b=0 and x2+px+q=0 have a common rootx2+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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