SOLUTION: given that the equation Z^2 + (p+iq)Z + r + is = 0 where p,q,r,s are real and non-zero has a real root then prove that pqs= s^2 + q^2r
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Question 485396: given that the equation Z^2 + (p+iq)Z + r + is = 0 where p,q,r,s are real and non-zero has a real root then prove that pqs= s^2 + q^2r
Answer by richard1234(7193) (Show Source): You can put this solution on YOUR website!
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:
 \pm \sqrt{(p+iq)^2 - 4(1)(r + is)}}{2})
 \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:
 \pm \sqrt{(p+iq)^2 - 4(r + is)} = k)
^2 - 4(r+is)} = k + (p + iq))
, square both sides.
^2 - 4(r + is) = k^2 + 2k(p+iq) + (p+iq)^2)
 = k^2 + 2k(p+iq))
Equating real and imaginary parts, we get
)
Replace k (we don't want k in our expression):
(-\frac{2s}{q} + 2p) = \frac{4s^2}{q^2} - \frac{4sp}{q})
Multiply both sides by q^2
, as desired.
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