What is the connection between the discriminant of a quadratic and the distance formula?

The $x$-coordinate of the center of a parabola $ax^2 + bx + c$ is $$-\frac{b}{2a}$$

If we look at the quadratic formula

$$\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}$$

we can see that it specifies two points at a certain offset from the center

$$-\frac{b}{2a} \pm \frac{\sqrt{b^2 - 4ac}}{2a}$$

This means that $\frac{\sqrt{b^2 - 4ac}}{2a}$ is the (horizontal) distance from the vertex to the roots. If I squint, the two squared-ish quantities being subtracted under a square root sign reminds me of the Euclidean distance formula

$$\sqrt{(x_0 - x_1)^2 + (y_0 - y_1)^2}$$

Is there a connection? If not, is there any intuitive or geometric reason why $\frac{\sqrt{b^2 - 4ac}}{2a}$ should be the horizontal distance from the vertex to the roots?

Solution 1:

Suppose that $a\ne 0$ and $b^2-4ac\geq 0$. Let $f(x)=ax^2+bx+c$ and $$ x_0= -\frac{b}{2a},\; x_1=\frac{-b+\sqrt{b^2-4ac}}{2a}, \;x_2=\frac{-b-\sqrt{b^2-4ac}}{2a}. $$ It follows that $$ y_0=f(x_0)=\frac{-b^2+4ac}{4a},\;y_1=f(x_1)=0, \; y_2=f(x_2)=0. $$ The distances from vertex to roots: $$ d_1=\sqrt{(x_1-x_0)^2+(y_1-y_0)^2}=\sqrt{\left(-\frac{b}{2a}-\frac{-b+\sqrt{b^2-4ac}}{2a}\right)^2+\left(\frac{-b^2+4ac}{4a}\right)^2}=\frac{\sqrt{(b^2-4ac)^2+4(b^2-4ac)}}{4|a|}>\frac{\sqrt{b^2-4ac}}{2|a|}, $$ $$ d_2=\sqrt{(x_2-x_0)^2+(y_2-y_0)^2}=\sqrt{\left(-\frac{b}{2a}-\frac{-b-\sqrt{b^2-4ac}}{2a}\right)^2+\left(\frac{-b^2+4ac}{4a}\right)^2}=\frac{\sqrt{(b^2-4ac)^2+4(b^2-4ac)}}{4|a|}>\frac{\sqrt{b^2-4ac}}{2|a|}. $$

Solution 2:

The connection you are looking for is simply Pythagoras theorem!

Imagine an upside down parabola. I ask that it be upside down so that the $y$ coordinate of the center is posiive. I do this just for convinience, my arguments are centainly true for any parabolas with two zeros.

You ask for a relationship between the equation for distance and the quadratic formula. Imagine two lines:

1) One connecting the center and the $x$-axis

2) One connecting the center and the zero.

The thing you have to see here, is that these two lines, when combined with the horizontal distance between the center and the zero, form a right triangle!

The length of line $1$ is simply the $y$ coordinate of the center, while the length of line $2$ (call it $H$) is given by the distance formula: $$H = \sqrt{(x_c-x_0)^2 + (y_c-y_0)^2}$$

And the horizontal distance is connected to these two by Pythagoras: $$y_c^2 + D^2 = H^2 \to D = \sqrt{H^2-y_c^2}$$

Your squinting was almost right. You just forgot that Pythagoras also gives these funny combinations of squares and square roots.

Let's do some algebra: The coordinates of the center are: $$x_c = \frac{-b}{2a} \quad \quad y_c = \frac{b^2(1-2a) + 2ac}{4a}$$

If we take the quadratic formula for granted, the coordinates of a zero are:

$$x_0 = \frac{-b}{2a} + \frac{\sqrt{b^2-4ac}}{2a} \quad \quad y_0 =0$$

We can calculate $H$, but since I'm lazy I'll calculate $H^2$: $$H^2 = (x_c-x_0)^2 + (y_c-y_0)^2$$ $$H^2 = \frac{b^2-4ac}{4a^2} + y_c^2$$

Note I left $y_c$ as is. You'll see why in a second:

The distance you want is:

$$D = \sqrt{H^2-y_c^2} = \sqrt{\frac{b^2-4ac}{4a^2} + y_c^2 - y_c^2} = \frac{\sqrt{b^2-4ac}}{2a}$$

So the link you were missing is the right triangle formed by the center, the zero and the $x$-axis. Note also that the distance formulate is derived from Pythagoras, so in reality the only connection is between Pythagoras and the quadratic formula.

Solution 3:

For the equation $ax^2 + bx + c = 0$ the second coefficients represents the average of the two roots:

$$ \frac{1}{2}(r_1 + r_2) = -\frac{b}{2a} $$

Instead of using the quadratic formula directly we can just observe that:

$$ (x - r_1)(x - r_2) = x^2 - (r_1 + r_2) + r_1 r_2$$

This also recovers the product mean of the two roots is $r_1 r_2 = \frac{c}{a}$.

In order to get the distance between the two roots we need to estimate the distance $|r_1 - r_2|$:

$$ |r_1 - r_2| = \sqrt{(r_1 + r_2)^2 - 4 r_1 r_2} = \sqrt{\frac{b^2}{a^2} - \frac{4c}{a}} = \frac{\sqrt{b^2 - 4ac}}{a}$$