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I am well aware of proofs of the quadratic formula that show, by completing the square and other methods, that the quadratic formula is what it is. I have been scouring the Internet and other resources I have and have not made a huge dent into why the quadratic formula works geometrically.

For example, from the geometry of the parabola, why is the axis of symmetry $ \frac{-b}{2a} $? (What does $ b $ mean geometrically? $ c $ represents the vertical shift from the origin and $ a $ refers to the "up or down pointing" of the parabola/maximum or minimum. I think it $ b $ has something to do with the vertex, hence completing the square.) I remember seeing something about this at one point, but I'm not sure where I put the paper I have.

Also, why is the distance from the axis of symmetry to the roots the square root of the discriminant divided by $ 2a $? Is there some geometrical reasoning for this that I cannot see?

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This isn't much more geometric, but consider that if the parabola $y = ax^2$ is moved to the right by $h$ and up by $k$, then the equation becomes $y = a(x-h)^2 + k$, which is just the same as $y = ax^2 - 2ahx + ah^2 + k$, so $h$ can be retrieved from $b$ by dividing by $-2a$.

Also, the roots are where $x - h = \pm\sqrt{-k/a}$, and since $k = c - ah^2 = c - b^2/4a$, it follows that $x - h = \pm\sqrt{b^2-4ac}/2a$.

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  • $\begingroup$ I guess I'm looking to see if there is a reason for dividing by $2a$. What's magic about that number from a visual standpoint? What's magic about $ b^2 - 4ac $ from a visual standpoint? That's the kind of thing I'm attempting to address. It seems that the numbers come from completing the square, but how do these numbers come up from a parabola? $\endgroup$
    – ZachTheRiah
    Commented Feb 3, 2015 at 12:26
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From Vieta's Formula, $\frac{-b}{a}$ is the sum of the two roots ($x$-intercepts). So, when we add the two $x$-intercepts and divide by $2$, we get half-way between them. That's going to be the $x$-coordinate of the vertex.

When it comes to the $\frac{\pm\sqrt{b^2-4ac}}{2a}$, that is the distance you move away from the center to get to the two $x$-intercepts.

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  • $\begingroup$ I like this perception: two numbers added result twice their mean. Moreover, both numbers have the same distance to that mean. How does this apply? The result as we all know is $(x-x_1)(x-x_2)$, what is $x^2-(x_1+x_2)x+x_1\cdot x_2$. Now set $x_1=m-n$ and $x_2=m+n$, thus $x_1+x_2=2m$. By comparing parts of $ax^2+bx+c$ it's easy to get $\displaystyle m=\frac{-b}{2a}$. Now use $x_1\cdot x_2=m^2-n^2$ to get $n$, and with a. m. $\displaystyle x_{1,2}=m\pm n$ "we are done". $\endgroup$
    – m-stgt
    Commented May 3 at 14:10

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