What are the roots of the following equation?

\[ x^2 + 5x + 6 =0 \]

If you're familiar enough with quadratic equations, then you might be able to tell right away that its roots are \(-2\) and \(-3\). If not, then you might at least have a vague sense that the left-hand expression can be factored into \( (x+2)(x+3) \). It seems obvious now that we have the factorization handy. Why? Well, we know that the roots interact with the coefficients of the equation such that \(2 \cdot 3 = 6\) and \(2+3 = 5\). More formally, if we have a polynomial that can be factored as \((x-r_1)(x-r_2)\) then we can expand it into:

\[ x^2 - (r_1 + r_2)x + r_1r_2 = a^2 - bx + c \]

\[ r_1 + r_2 = b\] \[ r_1r_2 = c \]

Now it's a little clearer why our example polynomial was so easy to factor. It's two roots must multiply to yield the last coefficient, \(c\), while they must add to give us the first coefficient, \(b\). The only way that can happen is with the roots \(2\) and \(3\). Let's try another one

\[ x^2 - x -1 \]

We run into a wall here. We can't really get away with guessing the solution now because there are no immediately obvious integer or rational solutions to this. In fact, there are none. So our task is to find a solution to the following nonlinear system.

\[ r_1 + r_2 = 1 = b\] \[ r_1r_2 = -1 = c \]

There's an interesting symmetry beneath these equations. In fact, we have no real way of distinguishing the two roots from each other. We can freely interchange \(r_1\) and \(r_2\) without changing the quantities in the above equations.

\[ r_1 \leftrightarrow r_2 \] \[ r_1 + r_2 = r_2 + r_1 = b\] \[ r_1r_2 = r_2r_1 = c \]

This is cool and all, but it's not helping us find the roots! We know that they must be different quantities from each other, but we're stuck at this point where we cannot distinguish them in any meaningful way.

Enter, Lagrange.

Lagrange's main insight concerned constructing non-symmetric permutations of these roots. If we're able to construct some intermediate equation that discriminates between the two roots, then we will have a way to solve this equation. Additionally, if we're trying to find the roots in terms of the coefficients of the polynomial, then we must write this equation in terms of the quantities \(r_1 + r_2\) and \(r_1 r_2\). We can start by going backwards; consider the equation:

\[ r_1 \not\leftrightarrow r_2 \] \[ r_1 - r_2 = q \not= r_2 - r_1 \]

Clearly, it doesn't have the same symmetry as the above equation; when we interchange the roots then the quantity changes from \(q\) to \(-q\). Since this distinguishes the roots from each other, this means that this quantity can be used to find solutions! Can we write this in terms of \(b\) and \(c\) though? Because this quantity changes sign when we permute the roots, we can square it in order to make it invariant under permutations of the roots.

\[ r_1 \leftrightarrow r_2 \] \[ (r_1 - r_2)^2 = r_1^2 + r_2^2 - 2r_1r_2 = q^2 \]

Now, let's see what happens when we square \(b\) and \(c\) too:

\[ b^2 = r_1^2 + r_2^2 + 2r_1r_2 \] \[ c^2 = r_1^2 r_2^2 \]

There won't be any terms involving \(c^2\), so we can just ignore it. We can rewrite the above equation for \(q^2\) as

\[ q^2 = r_1^2 + r_2^2 - 2r_1r_2 = b^2 - 4c \]

We can take the square root to get the difference between the roots. Now, we finally have an expression which is both in terms of the coefficients of the polynomial and also not symmetric with respect to its roots, which was what we were missing in our original system of equations.

\[ q = +\sqrt{b^2 - 4c} = r_1 - r_2 \]

Finally, we can plug this back into the original system of equations

\[ r_1 + r_2 = q + 2r_2 = b = 1\] \[ r_2 = \frac{1-q}{2} \; \; \; r_1 = \frac{1+q}{2} \; \; \; q = \sqrt{5}\]

Of course this is just our old friend the quadratic equation! You can check for yourself that these roots work the way we need them to.

While this can be seen as an alternative derivation of the quadratic equation, it's really more than that. Lagrange found this method while attempting to create a unified method of solving polynomial equations of all degrees - not just quadratics. In some senses, it worked! He uncovered some deep structures, such as the fact that taking the square root of a symmetric polynomial like \(q^2\) can sometimes lead to asymmetric polynomials which lead us to finding another polynomial's roots in terms of its coefficients. You could say that taking radicals effectively "splits" the roots into distinguished entities, foreshadowing the concept of splitting fields from Galois theory.

But in other senses, he failed in his goal. The method he developed had worked for cubics and quartics, but he eventually hit another wall when working with quintics - fifth-degree polynomials. It wasn't until later that Abel proved that there was no general equation that can solve fifth-degree polynomials, and it was even later until Galois was able to formulate why.

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## Comments

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TopNewestThis looks like math-made-unintentionally-hard but there's some more much richness behind it. I love how you introduced \(q,b,c\) here and yet it all loops back at the end! Thanks for sharing.

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Thank you! And theory-wise, it's extremely rich! This was historically one of the stepping stones to the development of group theory and abstract algebra.

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Group theory was around well before Lagrange; it was just never formalised properly. The origins of group theory lie in geometry, especially in the Greek times. What you are describing is a rough introduction to Galois theory, which is really just a formalisation of the "theory of equations". Also, Abel and Galois lived around the same time; only a couple of years separate their birth and death. Abel died of TB, whereas Galois was shot.

You should read up on the history of this stuff and check the accuracy of your statements.

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That looks very difficult to solve!

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It seems like the application of Vieta's formula. Viet Theorem. If quadratic equation ax2+bx+c=0 (reduced form is x2+ba+ca=0) has roots p and q, then p+q=−ba, pq=ca, i.e. sum of the roots equals the second coefficient, taken with opposite sign, and product of roots equals constant. Found it from here So I believe we can just use this long time ago proven theorem to solve equation quickly.

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@DYlAN MARSHALL

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