Analysis of the Complex Quadratic equation

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Analysis of some of the things explored on the previous page

We have az²+bz+c=0. The quadratic formula gives us the solutions

$\displaystyle z=\frac{{-b\pm \sqrt{{{{b}^{2}}-4ac}}}}{{2a}}$

If the discriminant is negative, we get

$\displaystyle \begin{array}{l}z=\frac{{-b\pm \sqrt{{-1\cdot \left( {4ac-{{b}^{2}}} \right)}}}}{{2a}}=\frac{{-b\pm \sqrt{{-1}}\sqrt{{4ac-{{b}^{2}}}}}}{{2a}}\\\ \ =\frac{{-b\pm i\sqrt{{4ac-{{b}^{2}}}}}}{{2a}}=-\frac{b}{{2a}}\pm \frac{{\sqrt{{4ac-{{b}^{2}}}}}}{{2a}}i=p+qi\end{array}$

So,

$\displaystyle p=-\frac{b}{{2a}},\quad q=\pm \frac{{\sqrt{{4ac-{{b}^{2}}}}}}{{2a}}$

Now, if we multiply q by 2a, we get

$\displaystyle 2aq=\pm \sqrt{{4ac-{{b}^{2}}}}$

Squaring both sides gives us

$\displaystyle 4{{a}^{2}}{{q}^{2}}=4ac-{{b}^{2}}$

This is an expression we will use in a couple of cases.

Case one, varying b

To explore this we might (seemingly strangely enough) eliminate b from the expression. In that way we get what will be invariant when changing b. We can use

$\displaystyle b=-2ap$

This gives us

$\displaystyle 4{{a}^{2}}{{q}^{2}}=4ac-4{{a}^{2}}{{p}^{2}}$

or

$\displaystyle {{a}^{2}}{{q}^{2}}=ac-{{a}^{2}}{{p}^{2}}$

Dividing through by a² we get

$\displaystyle {{q}^{2}}=\frac{c}{a}-{{p}^{2}}$

or

$\displaystyle {{p}^{2}}+{{q}^{2}}=\frac{c}{a}={{r}^{2}}$

In other words, a circle centred around the origin and with a radius of

$\displaystyle r=\sqrt{{\frac{c}{a}}}$

Case 2, varying a

We start in a similar way with

$\displaystyle 4{{a}^{2}}{{q}^{2}}=4ac-{{b}^{2}}$

From

$\displaystyle p=-\frac{b}{{2a}}$

we get

$\displaystyle a=-\frac{b}{{2p}}$

This gives us

$\displaystyle 4{{a}^{2}}{{q}^{2}}=4ac-4{{a}^{2}}{{p}^{2}}$

or

$\displaystyle {{a}^{2}}{{q}^{2}}=ac-{{a}^{2}}{{p}^{2}}$

Multiplying through by p² gives us

$\displaystyle {{b}^{2}}{{q}^{2}}=-2bcp-{{b}^{2}}{{p}^{2}}$

Then we divide by p² to get

$\displaystyle {{q}^{2}}=-\frac{{2cp}}{b}-{{p}^{2}}$

or

$\displaystyle {{q}^{2}}+{{p}^{2}}+\frac{{2cp}}{b}=0$

This will give us one of the rare cases when we actually need completing the square. We get

$\displaystyle {{q}^{2}}+{{p}^{2}}+\frac{{2cp}}{b}+{{\left( {\frac{c}{b}} \right)}^{2}}={{\left( {\frac{c}{b}} \right)}^{2}}$

or

$\displaystyle {{q}^{2}}+{{\left( {p-\frac{c}{b}} \right)}^{2}}={{\left( {\frac{c}{b}} \right)}^{2}}={{r}^{2}}$

I.e., a circle with the origin shifted by c/b, and the radius c/b. That would thus be a circle with the rim going through the origin.

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Last modified: Nov 16, 2025 @ 21:39