The year is 1646 and, as the nerd you have always been, you are reading a book about how to draw conic sections by the Dutch mathematician Frans van Schooten. However, in the back of the book there is an appendix looking at ways to solve cubic equations. This is strange.
In 2026 if I asked you to solve $x^3 = 13x + 12$, I imagine you could find a solution in just a few steps. But back in 1646 this is hard to do and mathematicians don’t like finding solutions that are negative or, God forbid, complex. In the book, Van Schooten actually describes negative solutions as false and complex solutions as fictitious.
In many ways the cubic equation is the open question of the century. While different approaches have been developed around the world, in 17th century Europe they have been battling with solutions to cubics for over a hundred years—in some cases literally (Google ‘Italian mathematics duel,’ and you won’t be disappointed). Various methods and solutions have been found for special cases, but there isn’t a simple way to find the roots of a general cubic.
So why would cubic equations show up in a book about conic sections?! It turns out that Van Schooten has found a way to represent a certain type of cubic equation using a geometrical drawing. By doing this, he is trying to bridge the gap between an exciting new branch of mathematics called ‘analytic geometry,’ cooked up by Descartes and Fermat, and the deductive geometry you have grown up with in the 1600s.
Van Schoot me a line
In 2026, you would have seen both types of geometry but probably would be far more comfortable using analytic geometry. Analytic geometry is about finding the equations of lines or circles and working out the information you need just using equations. You don’t need images or diagrams to help you. In contrast, deductive geometry relies on the fact that you have a diagram to help find your solution. You will need to construct circles, triangles and other shapes using a compass and a straight edge and then use your knowledge of angles and the properties of shapes to draw conclusions.
You may have come across deductive geometry at about the age of 14 or 15 in school where you were taught about circle theorems, and how to bisect a line or construct an equilateral triangle using a compass. I did not enjoy this topic in school. I struggled to understand why I had to use a compass and couldn’t just use my ruler to measure where the halfway point was on a line.
What my teacher didn’t say (or more probably my teenage self couldn’t comprehend) is that deductive geometry is incredibly useful and has been used by people across the world for thousands of years. It allows you to find an answer by building on what you already know. This is great if you want to find the new area of a field after the Nile floods in ancient Egypt or work out the size of container needed for a certain volume of crops.
Today, analytic geometry is taught and used much more heavily in mathematics classrooms than deductive geometry because it allows us to generalise. You don’t need to carry a compass and straight edge around everywhere you go because you can use equations to find your answer. You don’t need to draw a shape in the complex plane because you can use equations. Mathematicians love this and they often find deductive geometry hard because they just don’t have the practice!
However, I think it’s a shame not knowing what deductive geometry can do. The ideas in Van Schooten’s appendix are important because he is at the start of this change from deductive to analytic geometry. He draws together these two types of mathematics and tries to teach the reader how we can find a solution to a particular type of cubic equation.
So let’s solve a cubic equation like it’s 1646! We want to solve a cubic equation with a particular form: a depressed cubic. This means it looks something like $x^3 = px + q$. Here $x$ is our variable and $p$ and $q$ are positive real numbers. We’ll also throw in the extra condition that the square of half of the last term is less than the cube of one-third of the coefficient of the next to last term.
This is just a rather wordy way of saying that the cubic must satisfy $q^2/4 < p^3/27$. As an example, Van Schooten suggests the cubic equation $x^3 = 13x + 12$, which you can check satisfies the required properties.
Now, this seems pretty specific: no quadratic term, and that strange inequality involving $p$ and $q$. Actually, it’s not too bad: every cubic equation can be linked to a depressed cubic, so it’s only the inequality and the signs of $p$ and $q$ that hold us back.
So looking at our restriction of $q^2/4 < p^3/27$ you may immediately recognise the discriminant for a depressed cubic. You may also not. A depressed cubic (of the form $x^3-px-q = 0$) has as its discriminant the quantity $4p^3-27q^2.$ Like with a quadratic, looking at the sign of the discriminant tells us how many real roots the equation has. In our case, the requirement that $q^2/4 < p^3/27$ means the discriminant, $4p^3-27q^2$, is strictly positive. We should expect that the cubic will have three real roots (and later on we'll see that it does!).
Van Schoot your shot
Now’s the time to dig out that compass you haven’t used since you were 16, find a ruler (or straight edge), a piece of paper, and let’s get stuck in!
Let’s start by drawing a circle. Then pick two points, $F$ and $L$ on the circumference of the circle, and draw a line from $F$ to $L$, and another from $F$ going through the centre, $H$. Label the point opposite $F$ as $K$. Now mark another point, $I$, so that the distance around the circle from $L$ to $I$ is the same as the distance from $K$ to $L$.
Next, Van Schooten tells us to:
Join with straight lines the points KL, LI, and IF and continue the straight line FI outside of the circle to a point S, such that the angle FSL is equal to the angle IFL. From which SL is equal to LF, while SI is equal to FK.
That’s depicted above: the two line segments marked with blue dashes, $SL$ and $LF$, have the same length by construction, and so do the two segments shaded in green, $SI$ and $FK$.
It is less obvious that $SI$ must be the same length as $FK$, so let’s show this. Look at triangles $ILS$ and $KLF$. We know that $IL = LK$ because we have split the arc $IK$ into equal parts, and straight lines formed in this way are the same by proposition 29 in the third book of Euclid (in equal circles, equal circumferences are subtended by equal straight lines).
But also, as the angles $FKL$ and $FIL$ are opposite angles in a cyclic quadrilateral (highlighted in pink, below), they must add to 180 degrees, by proposition 22 in the 3rd book of Euclid.
So angle $LIS$ must also be the same as angle $LKF$, because $FS$ is a straight line! It may seem like a deep cut that Van Schooten is quoting Euclid but you have done the same! Any time you use a circle theorem to support your argument, or argue that a triangle is isosceles because two sides are the same length, then you are using Euclid. You can search for Euclid’s books online and you will see that they are full of these types of geometrical reasoning. Van Schooten just uses references like proposition 29, book 3 because this is how he has been taught.
Now we’ve got two triangles ($LFK$ and $ISL$) with two sides and one angle the same as each other. They must be two copies of the same triangle, which means we know that $FK$ must be the same length as $SI$, and the angles $LFK$, $LSI$, and $LFI$ must be the same as each other.
Next, Van Schooten tells us to add some more to our drawing. We add another point, $G$, to the circle, at the same distance from $I$ as $I$ was from $L$.
Then we continue:
To the same purpose extend FG to the point T until the angle FTI is equal to the angle GFI, in a similar manner TI equals IF, and TG themselves LF.
This is just repeating the logic we used a moment ago, with the triangle $FLS$: this time, it’s $LF$ and $TG$ that have the same length as each other (and also the same length as $SL$).
One more step:
Taking it further, the triangles FHL, FLS and FIT are similar: as before the ratio between HF and FL is the same as the ratio between LF and FS. Likewise for the same reason let the ratio between HF and FL the same as IF and FT.
In other words, the three shaded triangles (below) are similar, and so they all have the same ratios between their side lengths.
So we have all our triangles; it’s time to assign some values to each of the lengths.
Now, if you happen to be an ancient Greek mathematician, you will probably be saying “Elinor, Έλινορ, Elinor: we can’t assign lengths to the shapes we draw! If we do that we lose the abstractness, the essence of the shape!”
The ancient Greeks got really into doing maths in its ‘abstract form’. They wanted to build up all of geometry from only a few axioms. But being so restricted created certain problems for them. (The parallel postulate, squaring the circle and trisecting an angle may ring a bell.)
People had been using deductive geometry to find areas and volumes for thousands of years, and so of course they were using lengths and units. Although Van Schooten grew up learning this Greek mathematics, in this book he is linking deductive and analytic geometry. As such, by assigning lengths to part of the shape we are starting to think in terms of equations rather than diagrams.
Let’s keep following Van Schooten and set $HF = a$, $GF = b$, and $FL = x$.
Once we’re ready to choose them, $a$ and $b$ will be constants based on our equation, and we’ll try to work out the value of our variable $x$. Using our knowledge of the sine rule and similar triangles, we can then find a cubic equation from this diagram.
Firstly, we can conclude that $FS = x^2/a$ by using similar triangles. If the ratio between the long and short sides in triangle $FHL$ (the blue one, below) is $x/a$ and the short side in triangle $FLS$ (the pink one) has length $x$, then its long side has length $x^2/a$.
Next, we can work out the length $FI$. We know $$FS = \frac{x^2}{a},$$ and we worked out earlier that $SI$ was the same as $FK$, which we’ve now decided is equal to $2a$; so $$FI = \frac{x^2}{a}-2a$$:
Using the same ‘similar triangles’ logic, the long side in triangle $FIT$ must have length $$FT = \frac{x}{a} \left(\frac{x^2}{a}-2a\right) = \frac{x^3}{a^2}-2x.$$ Finally(!) we can work out the length of $FG$. Since $GT$ is the same length as $FL$ (in other words, length $x$), we know that, as shown below, $$FG = \frac{x^3}{a^2}-3x.$$
Now, finally, because we assigned $FG =b$, we have our cubic equation! It’s $b =x^3/a^2-3x$, which can be rearranged to $x^3 = 3a^2x + a^2b$. It even looks like the depressed cubics we met earlier, with $p=3a^2$ and $q=a^2b$.
We want to check that it’s of the correct form, that is, $$\frac{1}{4}q^2 < \frac{1}{27}p^3.$$
Substituting in the coefficients from our equation, we can see this as $a^4b^2/4<a^6$, which Van Schooten simplifies (dividing through and taking square roots) to $b<2a$.
This makes perfect sense in the context of our circle! We just need the line segment $FG$ (with length $b$) to be shorter than the circle diameter $FK$ (with length $2a$).
So we have constructed our diagram, assigned some lengths, and found a cubic equation. Great! Let’s now apply it to our particular equation, $x^3 = 13x + 12$.
Matching up $x^3 = 13x + 12$ with $x^3 = 3a^2 x + a^2b$, we need to have $a^2 = 13/3$, or in other words the radius of the circle should be $a = \sqrt{13/3}$.
We also conclude that $13b/3 = 12$, or $b =36/13$, and now all that’s left is to trisect an angle and work out what $x$ is. Easy. Cubic solved. Root found. Time to relax.
But there is one problem.
In 1646, we can’t trisect an angle. Remember earlier when I said that the Greeks struggled with some geometrical questions because they were very strict about what they could do with a ruler and compass? Well, they couldn’t trisect an angle. Mathematicians still hadn’t worked this out in 1646 and by 2026 we know that it is impossible with these strict rules. But all is not lost! Van Schooten embarks on another geometrical quest to find $x$ combining ideas from several different mathematicians. It takes him 17 pages (compared to the three pages of his book we’ve covered so far in this article), so I will spare you the details. Instead, here is another cool thing we can get out of the diagrams we’ve already drawn.
Van rooten
So far, we’ve only found one root of the cubic equation (that’s $x$). But, you might be thinking, surely there are two more? Where do they come into this? Once again, circles have the answers.
At the beginning of the appendix, Van Schooten tells us:
If you have an equilateral triangle MNL, where M, N, and L are all points on a circle and draw a line from L to a point F on the circumference in such a way that the line LF cuts MN at the point O and then draw lines MF, FN. We say that FL is equal to MF plus FN.
In the style of a university textbook, I will leave it to the reader to prove that the light and dark green lines have the same total lengths. One way to do it (the one in the book) involves spotting sets of similar triangles, which have the same ratios between the lengths of pairs of their sides.
For our specific example, $x^3 = 13x + 12$, we want each edge of the triangle to have length $\sqrt{13}$.
This side length of $\sqrt{13}$ might seem very random—but it means that the radius of the circle will be $\sqrt{13/3}$: our value of $a$, from earlier. (That’s another fact you can check: think about isosceles triangles that can be made from the radii and the edges of the original triangle.)
We can set off on another ‘similar triangles’ endeavour to show that $-FM$ and $-FN$ in the picture above are also roots of the equation! So if we can construct this triangle, we won’t just get one root of the cubic: we’ll get all three.
But we also know that the edges $FM$ and $FN$ added together make $FL$; that is, $FL = FM + FN$.
It turns out for a depressed cubic of this form, this will always be the case. This is because of an interesting relationship between the coefficients and roots of a cubic equation. If we take a general cubic of the form \[ ax^3 + bx^2 + cx + d = 0, \] with roots $\alpha, \beta, \gamma$ then $\alpha + \beta+ \gamma = -b/a$. The cubics we are looking at are all of the form $x^3 = px \pm q$. Because is no $x^2$ term, $-b/a = 0$ and so $\alpha + \beta + \gamma = 0$. This lets us conclude that the three roots of a cubic equation in this form will always sum to zero or, in other words, two of the roots sum to the negative of the third.
Van Schoot for the stars
You may think, “that was some cool maths but there is no way I would ever have come up with that approach to solving a cubic equation”. I would agree: at first I found it confusing and at points a little tedious going through every step. But I also haven’t been taught deductive geometry in the same way that Van Schooten was. I have been taught how to approach this type of question using analytic geometry and so of course I will use the method that seems easiest to me. Van Schooten is bringing together two different types of maths. We can see glimpses of the mathematics we are familiar with but there are also lots of different ideas.
This is what I find fascinating about the history of mathematics. You may not have come across this field before or realised that this is something that you can study. But it’s exactly what we’ve been doing here. We have looked into the past and followed the mathematics that somebody was doing in 1646! More generally, the history of maths looks at mathematicians from all over the world, and across time, to try to understand their work. We want to better understand different approaches to mathematics and different mathematical traditions. When you look at a range of methods and ideas, it helps you to better understand the underlying structure.
Next time you have a bit of time to spare, I encourage you to look up your favourite area of mathematics—or your favourite mathematician—to see what they were doing 380 years ago.
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