Prove that if $e=f+gi$ is a solution of the equation $az^2 + bz+c=0$ with $ a \in \mathbb R_0 $ and $ b,c \in \mathbb R $

Question

Prove that if $e= f+gi$ is a solution of the equation $ae^2 + be +c = 0$ with $a \in \mathbb R_0$; $b,c \in \mathbb R$ then the conjugate $\overline e= f-gi$ is also a solution of the equation.

Is it sufficient to use the discriminant formulae to get to the solutions as follows:

$$ y= \frac {-b+ \sqrt {b^2 -4ac}}{2a}$$ $$ y= \frac {-b- \sqrt {b^2 -4ac}}{2a}$$ ?

Answer 1

As you noted we know that the solutions of the equation are $$ e=\frac{-b}{2a} \pm \frac{\sqrt{b^2-4ac}}{2a} \qquad (1) $$ where $\frac{-b}{2a}$ is areal number

so, if $ e=f+ig$ we have: $\quad f=\frac{-b}{2a}$ and $\quad ig=\frac{\sqrt{b^2-4ac}}{2a}$ , but, from $(1)$ also $f-ig$ is a solution.

Answer 2

Hint: You have to use the rules for conjugate:

$$ \overline{z+w} = \overline{z}+ \overline{w}$$ $$ \overline{zw} = \overline{z}\cdot \overline{w}$$ $$ \overline{z} = z \iff z\in \mathbb{R}$$ and $$ \overline{z}^n = \overline{z^n}$$

Answer 3

Yes for quadratic case you have two solution $$\frac{-b}{2a} \pm \frac{\sqrt{b^2-4ac}}{2a}$$

and if complex, they are conjugate because the imaginary unit appears in front of the radical sign with $\pm $ signs which make them conjugates.

In general we have the same result that in case of real coefficients, complex roots of polynomials appear in conjugates.