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I noticed something that i find interesting about the quadratic formula. I hope someone can explain it to me.

$f(x) = Ax^2 + Bx + c$

$f'(x) = 2Ax + B$

If I say that $f'(x) = 0$ I get that $x = -\frac{B}{2A}$. From this I get that $f(-\frac{B}{2A})$ is a max/min point.

The quadratic formula is $x = \frac{-B \pm \sqrt{B^2-4AC}}{2A} = -\frac{B}{2A} \pm \frac{\sqrt{B^2-4AC}}{2A}$. Because $f(-\frac{B}{2A})$ is a max/min point this means that the distance between the max/min point to the points where $f(x)$ is $\frac{\sqrt{B^2-4AC}}{2A}$.

I tried to put$x = -\frac{B}{2A}$ into f(x) and got the following result:$f(-\frac{B}{2A})=A (-\frac{B}{2A})^{2} + B(-\frac{B}{2A}) + C = \frac{AB^2}{4A^2} - \frac{B^2}{2A} + C = \frac{B^2 -2B^2+4AC}{4A} = \frac{-B^2+4AC}{4A}$

If I take the root of it I get $\frac{\sqrt{4AC-B^2}}{2\sqrt{A}}$. It reminds me of the 2nd part of the quadratic formula$\frac{\sqrt{B^2-4AC}}{2A}$. Are they related? Why?

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  • $\begingroup$ $\;x=-\frac B{2A}\;$ is the well known $\;x-$ coordinate of the parabola's vertex, whereas $$f\left(-\frac B{2A}\right)=-\frac{\Delta}{4A}\;,\;\;\Delta=B^2-4AC$$ is the vertex's $\;y-$ coordinate $\endgroup$ – Timbuc Nov 6 '14 at 21:17
  • $\begingroup$ Ok, thank you. Do you know the relation between the vertex's $y$-cordinate and the distance between the vertex's $x$-cordinate and the $x$-points that make $f(x) = 0$? $\endgroup$ – peanut_butter Nov 6 '14 at 21:33
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Yes, you could say that they are very related!

In fact, you can develop an intuitive understanding for why these two expressions look so similar solely by resorting to the fundamental property of a quadratic (no calculus needed), namely that for a point a given x-length away from the vertex, you will be a y-length away from the vertex that is proportional to the square of the x-length.

To attain this fundamentally property, firstly note that an addition of a linear term or a constant to a quadratic only shifts the quadratic, it does not affect the curvature of the parabola (if you are unsure as to why, you could use calculus, though you could also figure this out by simply messing with quadratics). The only "thing" that actually affects the curvature of a quadratic is its "$a$" (x^2) term. So, if we are interested in studying the curvature of quadratics in general, we may as well just study the curvature of $y=ax^{2}$. Note that the vertex is (0,0). Now note:

As we move to a point of a given x-distance, we rise by a y-distance equal to (x-distance)^2 times $a$. I.e. $Δy=a(Δx)^{2}$

Hence, we have our fundamental property for quadratics.

To apply all this, first note that by finding $f(-B/2A)$, you have found the height of the quadratic at its vertex. By the quadratic formula, we know that to get into horizontal (x) alignment with either root, we need to shift horizontally by $(B^{2}-4AC)/2A$.

To get to the same y alignment as that of the roots (namely $y=0$), we clearly would have to move upwards/downwards by $f(-B/2A)$.

Now, we discussed that the fundamental property of a quadratic is that of the existence of a square relationship between x-changes from the vertex and respective y-changes, which does not rely on linear terms or constants! The vertex is the point from which the signature growth of a quadratic expresses itself. Hence, the square of the necessary x-distance, i.e. $(B^{2}-4AC)/2A$, to get to the roots must be equal to the necessary y-distance to get there multiplied by some constant ($1/a$). Or as you put it, the root of the necessary height is (almost) equal to necessary x-movement to get to a root.

By realizing that we are considering distances (as the necessary x/y-movement will always equal a positive value as we do not walk negative distances), we can take the absolute value of the root of $f(-B/2A)$ and the absolute value of $(B^{2}-4AC)/2A$. By considering the absolute value, we can equate $|B^{2}-4AC| = |4AC- B^{2}|$ (substitute) Now, note that the expressions are closer to being equal.

Still we have to multiply the x-movement by : $A^{0.5}$. This constant magically seems to make the 2 expressions equal, but the logic behind why it works is really simple (if you haven't already guessed). Note that we initially found out that the fundamental relationship between x-movements and y-movements in quadratics is of the form $y=ax^{2}$, so we have $y^{0.5}=a^{0.5}x$, hence why we need to multiply by a^{0.5} to gain equivalence.

Essentially, the similarity you noticed between the 2 expressions is related to noticing that a quadratic function expresses quadratic growth! I hope this answers your question. Feel free to ask for clarifications.

Have a nice day :).

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  • $\begingroup$ Just stating, I greatly edited my answer as some things in it were plainly wrong / convoluted. $\endgroup$ – Just_a_fool Nov 6 '14 at 22:27
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I follow you up to

$$f\left(-\frac{B}{2A}\right) = \frac{AB^2}{4A^2} - \frac{B^2}{2A} + C = \frac{4AC - B^2}{4A}.$$

But the square root of this gives you a $2\sqrt{A}$ in the denominator, so it's not quite what you have.

Here's the connection I see. In order to get real roots, $B^2 - 4AC$ needs to be non-negative.

We can rewrite the equation above as

$$f\left(-\frac{B}{2A}\right) = \frac{B^2 - 4AC}{-4A}.$$

We know $A \neq 0$. If $B^2 - 4AC = 0$, then $f(-B/2A) = 0$, which makes sense: a double real root.

If $B^2 - 4AC > 0$, then the sign of $A$ tells us the sign of $f(-B/2A).$ If $A$ is negative, then $f(-B/2A)$ is positive, and the parabola opens down to intersect the $x$-axis at the two roots. Likewise, if $A$ is positive, then $f(-B/2A)$ is negative, and the parabola opens up to intersect the $x$-axis at the two roots.

That might not be everything, but that's what I see.

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