A poster on the Google Plus Mathematics community commented that one feature of the Golden Ratio ϕ is that adding one to ϕ yields the same value as squaring ϕ does. That is, \[\phi^2 = \phi + 1\] He was surprised that there would be such a number.

While this is indeed an interesting attribute of ϕ, that led me to deeper musing about quadratics. There are actually two values with the property described above: ϕ and 1 – ϕ. We can see this by rewriting the original equation: Given a quadratic, the Fundamental Theorem of Algebra requires that there be two solutions, although they could be complex. So when looking for numbers that satisfy the equation \(x^2 = x + 1 \implies x^2 – x – 1 = 0\), we would expect there to be two values to satisfy it: We shouldn’t be surprised that such exist.

Nor should we be surprised that ϕ is the solution for a quadratic: Indeed, there are an infinite number of quadratics that have ϕ (or any other real value) as a solution. To see why, imagine a graph of a quadratic with one root pinned to a specific value. You can stretch it horizontally or vertically, infinitely.

So I asked myself: Is there a way to characterize the complete set of quadratics that have a root at a specific value? It seemed to me that there ought to be one.

The standard form of the quadratic is \(ax^2 + bx + c = 0\); its roots are \[\frac{-b \pm \sqrt{b^2 – 4ac}}{2a}\] I was already thinking about ϕ, that is, \(\frac{1 + \sqrt{5}}{2}\), which lends itself particularly well to the Quadratic Formula: If \(b = -a\), then ϕ will be a root when \(b^2 – 4ac = 5a^2\) (so that \(\sqrt{b^2 – 4ac} = a\sqrt{5}\)).

In such a case, we know that \(b^2 = a^2\), so \[b^2 – 4ac = a^2 – 4ac = 5a^2 \\ \implies -4ac = 4a^2 \implies c = -a\] This tells us that all quadratics of the form \(ax^2 – ax – a = 0\) will have a root at ϕ, which shouldn’t be a significant surprise, since \(ax^2 – ax – a = a(x^2 – x – 1)\).

This only accounts for quadratics where the right root is ϕ and the quadratic is vertically stretched. All of these quadratics have a left root of 1 – ϕ. What about cases where the left root is ϕ, or where the quadratic is horizontally stretched?

The algebra does get trickier at this point, so I’ll take the opportunity to generalize at the same time. Instead of ϕ, let us look for a formula for quadratics with a real root at some arbitrary value \(h\). I chose this variable name because, when graphed, this represents an arbitrary \(x\) value.

Let’s start by setting \(b = ka\). \(k\) is a fairly standard variable for an unknown coefficient, but it will be useful here because it’ll wind up being tied to the vertical position of the vertex of the parabola when the quadratic is graphed.

Our goal is to express \(c\) in terms of \(h\), \(k\), and \(a\). Replacing for \(b\) in the Quadratic Formula gives us: \[h = \frac{-ka \pm \sqrt{(ka)^2 – 4ac}}{2a}\]

At this point, we begin to solve for \(c\): \[2ah = -ka \pm \sqrt{(ka)^2 – 4ac} \\ \implies 2ah + ka = \pm \sqrt{(ka)^2 – 4ac} \\ \implies (2ah + ka)^2 = k^2a^2 – 4ac \]

Note that the last step above comes from squaring both sides, and so the plus/minus disappears. Swap the positions of the left hand side and the last term on the right, then expand the square: \[4ac = k^2a^2 – (2ah + ka)^2 \\ \implies 4ac = k^2a^2 – (4a^2h^2 + 4a^2kh + k^2a^2) \\ = -4a^2h(h + k) \\ \implies c = -ah(h + k)\]

This suggests for the general form: \[f(x) = ax^2 + kax – ah(h+k)\] For a given \(h\), where \(a\) is a non-zero real number, all and only quadratics of this or equivalent form will have a root at \(h\).

Let’s check this for ϕ, since this looks more complicated than \(f(x) = x^2 – x – 1\). In that case, \(a = 1\) and \(k = -1\). So does \(ah(h + k) = 1\)?

Well, \(h = \phi\), so \(ah(h + k) = \phi(\phi – 1) = \phi^2 – \phi\). But recall that \(phi^2 = \phi + 1\), so \(\phi^2 – \phi = \phi + 1 – \phi = 1\). Our general form works for the specific case.

What’s interesting about the general form is that it usually generates the same \(f(x)\) from two different \(h\) values. For instance, consider \(h = 2\), \(k = 1\), and \(a = 1\): \[f(x) = x^2 + x – 2(2+1) = x^2 + x – 6\] This function has two roots; the other is at -3. Sure enough, if we plug \(h = -3\) into the last term, we get \(3(-3 + 1) = 3(-2) = -6\).

So it’s equally capable of generating either the left or the right root. I say “usually” because some quadratics have their vertex on the x-axis, in which case there’s only a single real value for \(h\).

That’s also predictable, incidentally. The discriminant of the quadratic is \[b^2 – 4ac = (ka)^2 – 4(a)(-ah(h+k)) = k^2a^2 + 4a^2h(h+k) \\ = a^2(k^2 + 4h(h + k))\] If \(k = -2h\), then \[k^2 + 4h(h + k) = (-2h)^2 + 4h(h – 2h) \\ = 4h^2 + 4h^2 – 8h^2 = 0\] Thus, when \(k = -2h\), that is, when \[f(x) = ax^2 – 2hax + ah^2 \\ = a(x^2 – 2hx + h^2) = a(x – h)^2,\] the general formula only holds for a single value of \(h\).

This exploration started with the simple observation that \(\phi^2 = \phi + 1\) was based on the realization that \(x^2 = x + 1\) must be true for some value of \(x\), and ended with a general method of finding quadratics with a specific root. This sort of meandering is one of the hearts of mathematics.

JeremyWhy not just write the formula directly?

a(x-φ)(x-b) = 0

…then expand as needed.

Paul Hartzer(Post author)That’s fine, but it’s a different generalization.