Maths...complex~~~

px^2 + qx + r = 0
x = [-q + √(q^2 - 4pr)]/(2p) or [-q - √(q^2 - 4pr)]/(2p)

For x = [-q + √(q^2 - 4pr)]/(2p),
x
= [-q + √(q^2 - 4pr)] [-q - √(q^2 - 4pr)]/{2p[-q - √(q^2 - 4pr)]}
= (q^2 - q^2 + 4pr)/{2p[-q - √(q^2 - 4pr)]}
= 2r/[-q - √(q^2 - 4pr)]

Similarly, if x = [-q - √(q^2 - 4pr)]/(2p), then
x = 2r/[-q + √(q^2 - 4pr)]

If one of the roots is the reciprocal of the other, then
2r/[-q - √(q^2 - 4pr)] * 2r/[-q + √(q^2 - 4pr)] = 1
4r^2 = q^2 - q^2 + 4pr
r(p - r) = 0
r = 0 or p = r
As p, r are non-zero real numbers, so p = r.

step!!~~~~~~~~~~~~

px^2+qx+r=0

(i) x=[-q+/-√(q^2-4pr)]/(2p)
=2r/[-q+/-√(q^2-4pr)]

(ii) p=r