Now, the discriminant is essentially an algebraic quantity, while the determinant is essentially a geometric one. This suggests that there might be some interesting geometry hiding inside the problem of solving a quadratic equation-by which I mean more interesting than just finding where a parabola cuts the \(x\)-axis. And there is.

Let's make something new out of \(a,b\) and \(c\): the quadratic form \begin{equation*} Q(x,y) = ax^2+2bxy +cy^2 = \left( \begin{array}{cc} x&y \end{array} \right) \left( \begin{array}{cc} a&b\\b&c \end{array} \right) \left( \begin{array}{c} x\\y \end{array} \right) \end{equation*} Then a curve of the form \begin{equation*} Q(x,y) = k \end{equation*} is a conic section.

If you know a little linear algebra, then the next bit can be seen quite nicely in terms of the eigenvalues of the quadratic form. But if you already know that much linear algebra, you've probably already figured out what I'm about to explain, so I'm not going to take that route.

Fortunately, it is also possible to get there by a judicious application of some algebraic brute force, so that's what's about to happen.

We can find out what type the conic section is by completing the square on \(x\). \begin{equation*} \begin{split} Q(x,y) & = a(x^2+2\frac{b}{a}xy) + cy^2\\ &= a\left(x+\frac{b}{a}y\right)^2 + \frac{ac-b^2}{a}y^2\\ &= a\left(x+\frac{b}{a}y\right)^2 -\frac{1}{a} \Delta y^2 \end{split} \end{equation*}

Now there are three possibilities.

- If \(\Delta \lt 0\) then we have an ellipse (if \(a\) and \(k\) are the same sign);
- if \(\Delta=0\), a straight line; and
- if \(\Delta \gt 0\), a hyperbola.

- When \(\Delta \lt 0\) the only point on the curve is the origin, so there is no point \((x,1)\).
- When \(\Delta = 0\), the conic is the straight line \(x+by/a = 0\), so when \(y=1\) we have the single solution \(x=-b/a\).
- Finally, when \(\Delta \gt 0\), the conic is a pair of straight lines given by \begin{equation*} \sqrt{\Delta}y=\pm a \left( x+ \frac{b}{a} y\right) \end{equation*} and setting \(y=1\) gives us the usual solution.

So we can now see that the discriminant of the original quadratic equation is also (minus) the determinant of the quadratic form that describes a (degenerate) conic section, and solutions of the quadratic equation are where the line \(y=1\) intersects this conic. What previously gave us algebraic information (the presence or absence of real roots) now gives us geometric information. When the determinant is positive, the conic degenerates to the origin, with no solutions; when it is zero, to a single line, with one solution; and when it is negative, to a pair of lines and hence two solutions.

To summarize: we can think of the discriminant as an algebraic object, whose value determines how many (real) square roots it has and so how many solutions the quadratic has; or we can think of it as a geometric object, the determinant of a quadratic form, which determines how many intersection there are between a (degenerate) conic section and the line \(y=1\) and so how many solutions the quadratic has.

Addendum: Once upon a time, a mathematics student might expect to spend considerable time studying conic sections. It's a topic that seems to have largely dropped out of most syllabuses (both at school and university level), and I have a nagging suspicion that we're the poorer for it. Take a look at Salmon's conic sections if you don't believe me. Apart from containing a wealth of lovely mathematics, it's beautifully written (when did you last think that about a textbook?), and you may find it surprising to find what is explained and what is assumed as previous knowledge.

Thanks to David Bedford (@DavidB52s) for pointing out a typo in the calculation. It's gone now.

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