17 April 2025 · 2 min
Another way to derive the general solution to degree-two polynomials.
The symmetry of quadratic equations
In school, we learn to solve quadratic equations of the form using the discriminant. But we’re not always shown where the formula actually comes from.
Which is a shame, because the proof is well within everyone’s reach.
Let’s go
The classical method involves “completing the square”, which I’ve never found particularly natural. Instead, we can arrive at the same result using a simple observation about the symmetry of the equation, and a standard algebraic identity.
First, let’s simplify by dividing through by ( is non-zero, otherwise it’s not a quadratic equation):
Let and (think of them as scaled versions of and ), and separate the constant term:
The only awkward term here is the term. Without it, we could just take the square root. Factoring:
If only the left-hand side were of the form , the term would vanish…
A change of variable for clarity
Why not try a change of variable? Setting , so , we get:
This is a difference of squares, which reduces to:
We can now solve for and come back to afterwards.
Since :
And there’s the formula
Recalling that and :
Putting the contents of the square root over a common denominator:
Which gives us the general formula:
I give maths lessons online — if this kind of approach helps you understand things better, feel free to have a look here!