F.4 a.math Quadratic equation

上一步 :
-q(ar - cp) (cq - br) - p(cq - br)2 = r(ar - cp)2

抽common factor -(cq - br),即(br - cq)
得(br - cq) * [q(ar - cp) + p(cq - br)]
(br - cq)[qar - cpq + pcq - pbr] = r(cp - ar)2
(br - cq)[qar - cpq + pcq - pbr] = r(cp - ar)2 加減相消
(br - cq)[qar - pbr] = r(cp - ar)2 約掉 r
(br - cq)(aq - bp) = (cp - ar)2

If the quadratic equation ax^2+bx+c=0 and px^2+qx+r= 0 have one root in
common，prove that (br－cp)(aq－bp)=(cp－ar)^2
Sol
ax^2 + bx + c = 0 and px^2 + qx + r = 0 有一次公因數
=>消去2次式就是公因式
So
p(ax^2 + bx + c)－a( px^2 + qx + r)
=x(pb－aq)+(pc－ar)
Set x(pb－aq)+(pc－ar)=0
x=(pc－ar)/(aq－pb)
So
a[(pc－ar)/(aq－pb)]^2+b[(pc－ar)/(aq－pb)]+c=0
a(pc－ar)^2+b(pc－ar)(aq－pb)+c(aq－pb)^2=0
a(pc－ar)^2=－b(pc－ar)(aq－pb)－c(aq－pb)^2
=(aq－pb)[－b(pc－ar)－c(aq－pb)]
=(aq－pb)(－bpc+abr－acq+pbc)
=(aq－pb)(abr－acq)
=a(aq－pb)(br－cq)
So a(pc－ar)^2=a(aq－pb)(br－cq)
(br－cq)(aq－bp)=(cp－ar)^2