Question 485396
We can still use the quadratic formula to solve for Z. However we cannot simply set the discriminant to some number greater than 0. We instead solve for Z:


*[tex \LARGE Z = \frac{-(p+iq) \pm \sqrt{(p+iq)^2 - 4(1)(r + is)}}{2}]


*[tex \LARGE -(p+iq) \pm \sqrt{(p+iq)^2 - 4(r + is)} \in \mathbb{R}]


In other words, the numerator is a real number. Note that both roots do not necessarily have to be real; only one. We set the numerator equal to some real number k:


*[tex \LARGE -(p+iq) \pm \sqrt{(p+iq)^2 - 4(r + is)} = k]


*[tex \LARGE \pm \sqrt{(p+iq)^2 - 4(r+is)} = k + (p + iq)], square both sides.


*[tex \LARGE (p+iq)^2 - 4(r + is) = k^2 + 2k(p+iq) + (p+iq)^2]


*[tex \LARGE -4(r+is) = k^2 + 2k(p+iq)]


*[tex \LARGE -4r - 4si = k^2 + 2kp + 2kqi]


Equating real and imaginary parts, we get


*[tex \LARGE -4r = k(k+2p)]


*[tex \LARGE -2s = kq \Rightarrow k = -\frac{2s}{q}] Replace k (we don't want k in our expression):


*[tex \LARGE -4r = (-\frac{2s}{q})(-\frac{2s}{q} + 2p) = \frac{4s^2}{q^2} - \frac{4sp}{q}]


Multiply both sides by q^2


*[tex \LARGE -4q^2r = 4s^2 - 4pqs]


*[tex \LARGE 4pqs = 4s^2 + 4q^2r \Rightarrow pqs = s^2 + q^2r], as desired.