Consider az2+bz+c=0 where a,b,c are all complex numbers. What is the condition (ie the relation between a,b,c) for which the given quadratic has both real roots?

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Cecilia Nolan · Accepted Answer

I will assume that&amp;nbsp;a≠. Then, if&amp;nbsp;β=d,b―,a―&amp;nbsp;and&amp;nbsp;γ=d,c―,a―, the roots of&amp;nbsp;az2+bz+c&amp;nbsp;are the roots of&amp;nbsp;z2+βz+γ. If these roots are r,&amp;nbsp;s∈R, thenz2+βz+γ=(z−r)(z−s)&amp;nbsp;=z2−(r+s)z+rsand therefore&amp;nbsp;β=−(r+s)&amp;nbsp;and&amp;nbsp;γ=rs. In particular,&amp;nbsp;β,γ∈ℝ.Now, note that(r−s)2=(−(r+s))2−4rs&amp;nbsp;=β2−4γand therefore we must have&amp;nbsp;β2−4γ≥0. On the other hand, if indeed&amp;nbsp;β,γ∈ℝ&amp;nbsp;and if&amp;nbsp;β2−4γ≥0, it follows from the quadratic formula that the roots of&amp;nbsp;z2+βz+γ&amp;nbsp;are indeed real.So, the roots of&amp;nbsp;az2+bz+c&amp;nbsp;are real if and only ifd,b―,a―,d,c―,a―∈ℝ&amp;nbsp;and&amp;nbsp;(d,b―,a―)2−4d,c―,a―≥0.

Consider \(\displaystyle{a}{z}^{{2}}+{b}{z}+{c}={0}\) where a,b,c are all complex

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