We all know and love the quadratic formula. It says that the solutions to \(ax^2+bx+c=0\) are \[ x=\frac{-b\pm\sqrt{b^2-4ac}}{2a}. \]
However, high school algebra classes usually stop here. They do not, for instance, explain how one would solve an equation like \[x^3 + x + 1 = 0\] or \[x^4 - 6x^3 + 7x^2 + 8x = 9.\]
The goal of this series of two articles is to explain how to solve cubics (equations of the form \(ax^3 + bx^2 + cx + d = 0\)) and quartics (equations of the form \(ax^4 + bx^3 + cx^2 + dx + e = 0\)), like the ones above.
Though, really, these two articles have a different goal: we want to introduce ideas of Galois theory to the reader. Galois theory, invented by Évariste Galois, is an incredible branch of mathematics, which systematically studies the symmetry of solutions to equations. For example, the fact that the quadratic formula involves the funny \(\pm\) (plus or minus) symbol is related to a certain symmetry of quadratic equations. By thinking about the symmetries that cubics and quartics have, one is led to the formula for solving them.
Historically, Galois was born long after the cubic and quartic had been solved. He was interested in using symmetry to prove that you can go no further: there is no formula you can write down which solves equations of the form \[ax^5 + bx^4 + cx^3 + dx^2 + ex + f = 0\] using only radicals (things like \(\sqrt{5}, \sqrt[3]{2}, \sqrt[4]{7},\) etc.), addition, multiplication, division, and subtraction. Galois proved this by thinking about the kinds of symmetries that the solutions to a quintic have, and then showing no formula could ever have that same type of symmetry.
But enough with the preamble; let's get into solving some equations!
If we want to find a solution to the cubic or the quartic, it's useful to first ask how one might go about inventing the quadratic formula. And before we do that, it's worth asking: what does the quadratic formula even do, really?
The simplest sort of quadratic equation is something like \(x^2 = 17,\) where the solution is given by \(x = \pm\sqrt{17}\) (where, recall, the \(\pm\) means `plus or minus,' because both \(\sqrt{17} \approx 4.123\) and \(-\sqrt{17} \approx -4.123\) solve the equation \(x^2 = 17\)).
But something was hidden when we wrote \(\sqrt{17} \approx 4.123.\) The symbol \(\sqrt{17}\) just means "the (positive) number which squares to \(17.\)" So, when you ask someone for a solution to \(x^2 = 17,\) and they answer back "it's \(\pm\sqrt{17}\)", in some sense they've done nothing for you. The only reason we consider \(\sqrt{17}\) as being a valid answer to \(x^2 = 17\) is because the square root function comes up very often, and so humanity has developed many tools for quickly calculating it -- for example, your calculator probably has a button for it. People thus consider \(\sqrt{17}\) to be a valid expression to use in equations, and leave the task of numerically evaluating it to the reader of their solution.
Similarly, when one writes that \[ax^2 + bx + c = 0\] has solutions \[x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a},\] one is essentially saying that if you know how to solve the equation \[y^2 = b^2 - 4ac,\] then you can solve \(ax^2 + bx + c = 0\) as well.
So, the quadratic formula is secretly a tool for turning one quadratic into a simpler quadratic. This reduction is more useful than it might seem, though: there are tons of algorithms for computing square roots (your calculator knows one, your computer knows one, and you could even learn one to execute with pen-and-paper for yourself!). So, the quadratic formula reduces an arbitrary quadratic equation to one for which we already have a solution.
People often joke that mathematicians are lazy, because this is incredibly common in mathematics: if you have some problem you cannot solve, like \(-x^2 - 2x + 1 = 0,\) then reduce it to some problem you can solve, like \(y^2 = 8.\) The quadratic formula is just a reduction of this type. Human science and engineering is built on the work of giants: if the people before you have already discovered something, it is a good idea to try and reframe your problem in a way that their past discoveries are applicable.
We now have some understanding of what the quadratic formula is: it's a tool for turning quadratic equations into square root problems. But how does one actually invent the quadratic formula? For this, we turn to an idea of Vieta.
Suppose the quadratic equation \[x^2 + bx + c = 0\] has the two solutions \(x=\alpha\) and \(x=\beta.\) This means we can factor the quadratic as \[x^2 + bx + c = (x-\alpha)(x-\beta).\]
The French mathematician Vieta noticed something amusing. If we expand out \((x-\alpha)(x-\beta),\) we get \[(x-\alpha)(x-\beta) = x^2 - (\alpha + \beta)x + \alpha\beta.\] Thus, if \[x^2 + bx + c = (x-\alpha)(x-\beta),\] then \[\alpha + \beta = -b\] and \[\alpha\beta = c.\]
These are known as Vieta's formulas, which relates the solutions \(\alpha\) and \(\beta\) of the quadratic to its coefficients \(b\) and \(c\).
Vieta's formulas mean that, even if you haven't yet solved your quadratic equation, you can deduce certain things about the solutions. For example, just by looking at the equation \[x^2 + bx + c = 0,\] I can tell that the two solutions multiply to get \(c,\) even without solving the equation.
If you're a bit more clever, you can also figure out more information about the solutions \(\alpha\) and \(\beta\) to the above equation without actually solving it.
For example, we secretly know the value of \(\alpha^2+\beta^2\), since \[ \alpha^2+\beta^2=(\alpha+\beta)^2-2\alpha\beta= (-b)^2 - 2 \cdot c = b^2 - 2c \]
But what about \(\alpha+2\beta\), or \(\alpha-3\beta^2\)? Is there any chance that we could secretly evaluate them without actually solving our equation?
Notice that the expressions \(\alpha+\beta\), \(\alpha\beta\), and \(\alpha^2+\beta^2\) are all symmetric: they don't change when we swap the value of \(\alpha\) and \(\beta\). That \(\alpha+\beta\) and \(\alpha\beta\) are symmetric shouldn't come as a surprise, since the expression \((x-\alpha)(x-\beta)\) is already symmetric in \(\alpha\) and \(\beta\).
In fact, any symmetric polynomial expression of \(\alpha\) and \(\beta\) can be rewritten as a polynomial in terms of \(b\) and \(c.\) For example, \(\alpha^3 + \beta^3\) is symmetric in \(\alpha\) and \(\beta,\) and we can write \[\alpha^3 + \beta^3 = (\alpha + \beta)^3 - 3(\alpha+\beta)\alpha\beta = -b^3 + 3bc.\]
How is all of this useful in finding the quadratic formula? The key insight is that, to solve for \(\alpha\) and \(\beta\), we eventually need to break the symmetry.
As soon as you find one quantity which is not symmetric, you can use it to solve for \(\alpha\) and \(\beta\) separately. For example, suppose you were trying to solve \[x^2 + x -6 = 0,\] and I told you that the solutions \(\alpha, \beta\) obeyed \(\alpha - \beta = 5.\) Then you could solve this quadratic without much work, because you can deduce just by looking at its coefficients that \(\alpha + \beta = -b = -1,\) and so \[\alpha + \beta = -1,\] \[\alpha - \beta = 5.\] This is a system of equations which you can easily solve to get \(\alpha = 2, \beta = -3.\)
The expression \(\alpha - \beta\) is not symmetric in \(\alpha\) and \(\beta\); if you switch the values of \(\alpha\) and \(\beta,\) then \(\alpha-\beta\) gets multiplied by \(-1.\) This is why being told the value of \(\alpha-\beta\) it makes it so easy to solve the quadratic: as soon as you can break the symmetry, you can solve the quadratic.
So, how we can determine \(\alpha-\beta\)? Notice that it's very close to being symmetric. In fact, its square \[ (\alpha-\beta)^2 \] will really be symmetric: swapping \(\alpha\) and \(\beta\) multiplies the value of \(\alpha-\beta\) by \(-1,\) but \((-1)^2 = 1,\) so swapping \(\alpha\) and \(\beta\) won't change \((\alpha-\beta)^2\) at all!
And indeed, we can express \((\alpha-\beta)^2\) as a polynomial of \(b\) and \(c\): \[ (\alpha-\beta)^2=(\alpha+\beta)^2-4\alpha\beta=b^2-4c. \] This tells us \[ \alpha-\beta=\pm\sqrt{b^2-4c}.\] Combining with \(\alpha+\beta=-b\), we can solve for \(\alpha\) and \(\beta\): \[ \alpha,\beta=\frac{-b\pm\sqrt{b^2-4c}}2. \]
This is the familiar quadratic formula! You can derive the full formula by replacing \[ax^2 + bx + c = 0\] by the equation \[x^2 + \frac{b}{a}x + \frac{c}{a} = 0.\]
What happens if we try solving more complicated equations with Vieta's formulas? To see, let's start with a cubic equation \[ x^3+ax+bx+c=0. \]
Label the solutions \(x=\alpha,\beta,\gamma\). If we expand out the expression \[x^3 + ax + bx + c = (x-\alpha)(x-\beta)(x-\gamma),\] then we can find that \[ a=-(\alpha+\beta+\gamma), \] \[b=\alpha\beta+\alpha\gamma+\beta\gamma,\] and \[c=-\alpha\beta\gamma. \]
Notice how the quantities \(a, b,\) and \(c\) are all symmetric functions of \(\alpha,\beta,\) and \(\gamma.\) And, conversely, if we have a polynomial symmetric in \(\alpha,\beta,\gamma,\) we can always rewrite it in terms of \(a,b,c.\) For example, \(\alpha^2+\beta^2+\gamma^2\) is symmetric, and indeed you can check using Vieta's formulas that \[ \alpha^2+\beta^2+\gamma^2=a^2-2b. \] This means that, just by looking at our cubic equation, we can deduce the value of \(\alpha^2 + \beta^2 + \gamma^2,\) without needing to solve for the numbers \(\alpha,\beta,\gamma\) individually.
To solve the cubic equation, we need to break the symmetry somehow, just as we did for the quadratic equation. This is a bit trickier than breaking the symmetry for the quadratic equation; next week, we'll explain a more systematic way of looking at the symmetries of \(\alpha,\beta,\) and \(\gamma,\) which will help us solve the cubic. But for now, we're going to explain how to solve quartic equations, assuming we already know how to solve cubics.
In 1540, the Italian mathematician Lodovico de Ferrari discovered something remarkable: just as one can use the quadratic formula to reduce solving equations \[ax^2 + bx + c = 0\] to solving the simpler equations \[y^2 = d,\] one can similarly use a certain clever algebra trick to reduce solving quartic equations \[ax^4 + bx^3 + cx^2 + dx + e = 0\] to cubic equations \[fy^3 + gy^2 + hy + i = 0.\]
At the moment, this reduction won't be super helpful to us, because we haven't solved the cubic yet -- that will wait for next week. But actually, we're following the history quite well, because when Ferrari first discovered this trick, he also didn't know how to solve the cubic equation! Ferrari's teacher, the Italian mathematician Gerolamo Cardano, later learned of a method for solving cubic equations, so that Ferrari's trick allowed one to solve quartics as well.
Now, onwards to Ferrari's idea. Let's think about the quartic equation \[ x^4+ax^3+bx^2+cx+d=0, \] and denote the solutions as \(\alpha,\beta,\gamma,\delta\). Our hope is to levarage the symmetry of the solutions to reduce this to a cubic equation.
A cubic polynomial has three solutions. Our goal, then, is to write down some simple cubic polynomial \[(x-r)(x-s)(x-t)\] which has the following two properties: the coefficients of \((x-r)(x-s)(x-t)\) should be expressible in terms of \(a,b,c,\) and \(d\) -- equivalently, the coefficients should be symmetric functions of \(\alpha,\beta,\gamma,\) and \(\delta\); and \(r, s,\) and \(t\) some non-symmetric functions of \(\alpha,\beta,\gamma,\delta,\) so that if we learn the values of \(r, s, t\), then we can break the symmetry and solve the quartic.
These two goals are somewhat at odds with each other: we want the coefficients of \((x-r)(x-s)(x-t)\) to be symmetric in the roots, but we want the roots \(r,s,t\) to each, individually, be non-symmetric in \(\alpha,\beta,\gamma,\delta.\)
Ferrari observed that we could take \[ r = \alpha\beta+\gamma\delta,\] \[s = \alpha\gamma+\beta\delta,\] and \[t= \alpha\delta+\beta\gamma. \]
Individually, \(r, s, t\) are not symmetric in \(\alpha,\beta,\gamma,\delta.\) But together, they have a common symmetry; for example, if we swap the values of \(\alpha\) and \(\beta,\) then \(r\) stays the same (because \(\alpha\beta\) becomes \(\beta\alpha\), which is unchanged), but \(s\) and \(t\) get swapped. Similarly, if we swap \(\alpha\) and \(\gamma,\) then \(s\) stays the same, but \(r\) and \(t\) get swapped. Generally, permuting the values of \(\alpha,\beta,\gamma,\delta\) just permutes the quantities \(r,s,t.\)
That means the coefficients of the polynomial \[ \big(x-(\alpha\beta+\gamma\delta)\big)\big(x-(\alpha\gamma+\beta\delta)\big)\big(x-(\alpha\delta+\beta\gamma)\big) \] are symmetric functions of \(\alpha,\beta,\gamma,\) and \(\delta.\) For example, this cubic equation has \(x^2\) coefficient \[-r-s-t = -\alpha\beta - \gamma\delta - \alpha\gamma - \beta\delta - \alpha\delta - \beta\gamma,\] which is symmetric.
That means we know the coefficients of this cubic \((x-r)(x-s)(x-t)\) in terms of the original coefficients \(a,b,c,d.\) We call the result the resolvent cubic, and with some tedious but doable algebra, one can work out that \[(x-r)(x-s)(x-t) = x^3 - bx^2 + (ac - 4d)x + (4bd - c^2 - a^2d).\]
If we knew how to solve the cubic (which we will next week!), then we'd be able to solve it to figure out the values of \(r=\alpha\beta+\gamma\delta\), \(s=\alpha\gamma+\beta\delta\), and \(t=\alpha\delta+\beta\gamma\).
Now, how do we write \(\alpha,\beta,\gamma,\delta\) in terms of the numbers \(r,s,t\)? To start, observe that \[r = (\alpha\beta) + (\gamma\delta),\] and \(d = (\alpha\beta)(\gamma\delta)\) (where \(d\) was the constant term of our quartic). Thus, we have two numbers \(\alpha\beta\) and \(\gamma\delta,\) and we know their sum \(r\) and their product \(d.\) As we saw when looking at Vieta's formulas for quadratics, figuring out two numbers given their sum and their product is equivalent to solving a quadratic equation -- and in particular, we can use the quadratic formula to solve for \(\alpha\beta\) and \(\gamma\delta\).
Similarly, we can use the same trick with \(s\) and \(t\) to compute the values of \(\alpha\gamma, \alpha\delta, \beta\gamma,\) and \(\beta\delta.\)
Next, observe that \[ r+s=\alpha\beta+\gamma\delta+\alpha\gamma+\beta\delta=(\alpha+\delta)(\beta+\gamma). \] We also know \[ (\alpha+\delta)+(\beta+\gamma)=-a, \] so by solving the relevant quadratic equation again we can compute \(\alpha+\delta\) and \(\beta+\gamma\). Since we already solved for \(\alpha\delta\) and \(\beta\gamma\) above, we again solve the relevant quadratic equations to compute all four solutions \(\alpha,\beta,\gamma,\delta\).This lets us solve quartic equations to our heart's desire (or at least it will, once we find the cubic formula...), but writing out the full formula in terms of the variables \(a,b,c,d\) will take ages! The formula is incredibly long; you can view it on Wikipedia if you so desire.
You might find it strange that, at no point in this derivation did we ever explicitly compute a fourth root like \(\sqrt[4]{6}.\) Observe, however, that we did need to repeatedly apply the quadratic formula, which uses square roots; as \[\sqrt[4]{6} = \sqrt{\sqrt{6}},\] our iterated applications of the quadratic formula are secretly the same as taking fourth roots.
We end this section with a remark for experts in group theory: secretly, the 'exceptional homomorphism' \(S_4 \to S_3\) of symmetric groups was responsible for the resolvent cubic. Can you see why?
For a long time, people unsuccessfully tried to extend Cardano and Ferrari's method to solve quintic equations. The problem remained unsolved for over 300 years, but in the 19th century Galois (our friend from the introduction!) proved that there doesn't exist a general formula for the solutions to quintic equations and beyond!
As we remarked in the introduction, Galois proved this by considering the symmetries of quintics, and showing that no formula could have the right symmetries to give quintics. We hope that today's article with Vieta's formulas, and next week's article with the cubic formula, will help you understand what it means for solutions of an equation to have symmetry.
As a historical aside, we remark that Galois had a quite tumultuous life. He invented Galois theory as a teenager (and went quite far with it!), but was rejected from university (apparently, at an oral exam given to determine who would be admitted, he tried to explain Galois theory to the examiner instead of answering their questions).
Galois tragically died at the age of 20, after being shot and killed in a duel. His work might have been forgotten, had he not spent the night before the duel writing a letter containing his ideas to his friend Chevallier.
While it took around a decade, eventually the mathematical world realized how important Galois' work was. The famous mathematician Hermann Weyl wrote that "[Galois'] letter, if judged by the novelty and profundity of ideas it contains, is perhaps the most substantial piece of writing in the whole literature of mankind."
