Quadratic Equation 2

Let α be the common root, then
α2 + aα + b = 0 = α2 + pα + q
b - q = α(p - a)
α = (b - q)/(p - a)
Then sub it back into the original equation:
α2 + aα + b = 0
(b - q)2/(p - a)2 + a(b - q)/(p - a) + b = 0
(b - q)2 + a(b - q)(p - a) + b(p - a)2 = 0
(b - q)2 + (p - a)[a(b - q) + b(p - a)] = 0
(b - q)2 + (p - a)[ab - aq + bp - ba] = 0
(b - q)2 + (p - a)(- aq + bp) = 0
(b - q)2 + (p - a)(bp - aq) = 0
(b - q)2 = (a - p)(bp - aq)

Let the common root be k. Then
k^2 + ak + b = 0 ..........(1)
k^2 + pk + q = 0...........(2)
(1) - (2) we get
ak + b - pk - q = 0
k(a - p) = q - b
k = (q-b)/(a-p)..............(3)
p x (1) - a x (2) we get,
pk^2 + apk + bp - ak^2 - apk - aq = 0
k^2( p - a) = aq - bp
k^2 = (aq- bp)/(p -a).........(4)
So the square of (3) = (4), that is
(q-b)^2/(a-p)^2 = (aq- bp)/(p -a)
So (q-b)^2 = (aq- bp)(p -a).