f.4 a maths 一問

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