Solving quadratic equations - Lagrange style

What are the roots of the following equation?

x2+5x+6=0 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 2-2 and 3-3. If not, then you might at least have a vague sense that the left-hand expression can be factored into (x+2)(x+3) (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 23=62 \cdot 3 = 6 and 2+3=52+3 = 5. More formally, if we have a polynomial that can be factored as (xr1)(xr2)(x-r_1)(x-r_2) then we can expand it into:

x2(r1+r2)x+r1r2=a2bx+c x^2 - (r_1 + r_2)x + r_1r_2 = a^2 - bx + c

r1+r2=b r_1 + r_2 = b r1r2=c 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, cc, while they must add to give us the first coefficient, bb. The only way that can happen is with the roots 22 and 33. Let's try another one

x2x1 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.

r1+r2=1=b r_1 + r_2 = 1 = b r1r2=1=c 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 r1r_1 and r2r_2 without changing the quantities in the above equations.

r1r2 r_1 \leftrightarrow r_2 r1+r2=r2+r1=b r_1 + r_2 = r_2 + r_1 = b r1r2=r2r1=c 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 r1+r2r_1 + r_2 and r1r2r_1 r_2. We can start by going backwards; consider the equation:

r1↮r2 r_1 \not\leftrightarrow r_2 r1r2=qr2r1 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 qq to q-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 bb and cc 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.

r1r2 r_1 \leftrightarrow r_2 (r1r2)2=r12+r222r1r2=q2 (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 bb and cc too:

b2=r12+r22+2r1r2 b^2 = r_1^2 + r_2^2 + 2r_1r_2 c2=r12r22 c^2 = r_1^2 r_2^2

There won't be any terms involving c2c^2, so we can just ignore it. We can rewrite the above equation for q2q^2 as

q2=r12+r222r1r2=b24c 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=+b24c=r1r2 q = +\sqrt{b^2 - 4c} = r_1 - r_2

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

r1+r2=q+2r2=b=1 r_1 + r_2 = q + 2r_2 = b = 1 r2=1q2      r1=1+q2      q=5 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 q2q^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.

Note by Levi Walker
1 year ago

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

Pi Han Goh - 1 year ago

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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.

Levi Walker - 1 year ago

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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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@A Former Brilliant Member I know it was around, but only implicitly. Galois was the first to formalize the notion of a group during his work on quintic equations. Before that, Lagrange created the resolvent method which Abel picked up to prove that quintic equations have no general solution. The operations and relationship between the resolvents was essentially group theory before its formal development, but Galois was the one who first described it in abstract terms.

Levi Walker - 1 year ago

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@Levi Walker Yup! Thanks for clarifying. You should write something up regarding quintics.

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@A Former Brilliant Member I think I will as a followup - I might even do a wiki page about Galois theory, too, but I'm not sure yet.

Levi Walker - 1 year ago

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That looks very difficult to solve!

Anton Goetze - 1 year ago

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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.

Scarlett Bolton - 8 months ago

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