Polynomials: Px^2+qx+r=0

Why does 1/α^2 + 1/β^2 equal tol q^2=2pr)/r^2
if (alpha) and (beta) are the roots px^2+qx+r=0
alpha+beta=-q/p alpahbeta=r/p

I know we eventually get this by subbing
=[(β+α)-2αβ]/(α+β)^2
Which equal to
[(q^2)/(p)-(2r)/(p)]/(r^2/p^2) --> providing (r^2/p^2) = alphabeta squared
---This is where i am confused do we flip αβ? and then expand to numerator??

1/a^2 + 1/b^2 = (a^2+b^2)/(ab)^2 = ((a+b)^2 - 2ab)/(ab)^2
since you know a+b = -q/p, and ab = r/p
just sub in and you should get
(q^2/p^2 - 2r/p) / (r^2/p^2)
after some simplification you should get
(q^2 - 2pr) / r^2 or (q^2/r^2 - 2p/r)

α and β are the roots of : px^2+qx+r = 0
x^2 +(q/p)x +r/p = 0 ........................................… [1]
SO,
x^2 - (α+β)x + αβ = 0 ........................................… [2]

Comparing co-efficients of [1] and [2], we get,
(α+β) =-q/p ........................................… [3]
and
αβ = r/p ........................................… [4]
α = r/pβ ...................................... [5]

From [3] and [5], r/pβ + β = -q/p
r/p + β^2 = -qβ/p, ==> β^2 +qβ/p +r/p = 0
β = [-q/p +/- sqrt(q^2/p^2 - 4rq/p^2)] / 2

β = [-q +/- sqrt(q^2 -3rq)] / 2p
Thus,
1/α^2 + 1/β^2 = (q^2 + 2pr) / r^2 >==========< ANSWER