Proof without words of the Quadratic Formula?

Question

As suggested by @Moti and @YvesDaoust in this post, a simple way to identify the roots (red dots) of a parabola (given focus and directrix, blue) by means of straightedge and compass is to draw the circle with center in the focus and radius the distance $\overline{HI}$ between the $x$-axis and the directrix.

Now, consider the Quadratic Formula

$$ \color{red}{x_{\pm}}=\frac{-b\pm\sqrt{b^2-4a \cdot \mathbf{c} }}{2a}. $$

In the above image (a part the roots), it is easy to spot the term $\mathbf{c}$, i.e. the intercept of the parabola with the $y$-axis.

My question is:

How to geometrically illustrate the other various algebraic terms of the quadratic formula by means of this construction, is such a way that the algebraic relation results immediately evident?

With geometrically, I mean some visual intuition, based on such plot (or something similar), in the spirit of a "proof without words".

Thanks for your help!

Answer 1

Here's a slight re-packaging of notions from my previous answer.

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$$|OQ_{\pm}| \;=\; |BB_{-}| \pm |MQ_{+}| \;=\;-\frac{b}{2a} \pm \sqrt{\frac{b^2}{4a^2}-\frac{c}{a}} \;=\; \frac{1}{2a} \left(\;-b \pm \sqrt{b^2-4ac}\;\right)$$

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• The figure represents the scenario in which $a>0$, $b\leq 0$, $c\geq 0$ (and thus that $|OM|\geq |MQ_{\pm}|$). Adjustments to accommodate various sign changes should be clear.

• Figure labels and calculations incorporate the fact that the latus rectum has length $1/a$.

• That $\overline{OM} \cong \overline{BB_{-}}$ is my previous answer's Property 2. That these segments' common signed length is $-b/(2a)$ follows, as before, from the equation of the represented parabola, by calculating the difference in $y$-coordinates for points with $x$-coordinates $\pm 1/(4a)$.

• The relation between the highlighted areas follows immediately from my previous answer's Property 1. Stripping away the trappings of the specific problem, we can state the area property as a general principle that @Andrea should appreciate:

Property 1a. If squares are erected upon axis-perpendicular semi-chords of a parabola, then the difference in their areas is the area of the rectangle bounded by those chords and the extremities of the parabola's latus rectum.

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By the way, here's a proof-without-words for my previous question's Property 2, using Property 1a above.

Answer 2

As the Wikipedia article Power of a point indicates, the tangent to the circle from the origin distance squared is the product of the two roots, but this is just $\,c/a.$

Answer 3

This solution isn't nearly as self-evident as I like my illustrations to be, but there are some interesting ideas here.

I'll preface this by noting, in something of an echo of @Rahul's comment, that geometricizing $y=ax^2+bx+c$ is a little tricky, in that $a$, $b$, $c$ are dimensionally distinct. In the approach described below, we take $x$ and $y$ (and thus also the roots of the quadratic equation) to be represented by ($1$-dimensional) lengths; necessarily, we see that $c$ must also be $1$-dimensional, $b$ must be $0$-dimensional (a ratio), and $a$ must be ... $(-1)$-dimensional!

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Suppose that the graph of $y=ax^2+bx+c$ represents an upward-facing parabola with vertex $V= (h,-k)$; that is, we take $a$ positive and $b$ non-positive. Let $f$ be the vertex-to-focus distance, $f := |VF|$. Let the parabola cross the $y$-axis at $C$, of distance $c\geq0$ from the origin (although it's less problematic here to allow $c<0$), and let the parabola cross the $x$-axis at $R_{\pm}$, at distances $h\pm s$ from the origin.

Some auxiliary points: Let the $x$-axis and parabola axis meet at $M$ (the midpoint of $R_{+}$ and $R_{-}$). Let the horizontal line through $V$ meet the $x$-axis at $k$, and let $S$ be the projection of $R_{+}$ onto that line (so $|VK|=h$, $|VS|=s$, and $|OK|=|SR_{+}|=k$). Also, let the lines $x=\pm f$ meet the parabola at $B_{\pm}$, and let $B$ complete the right triangle with hypotenuse joining those points. Points $A$ and $G$ are on the $y$-axis and parabola axis such that $|KA|=|VG|=4f$.

Given the above, the below happens to be an illustration of the Quadratic Formula:

As I mentioned: not nearly as self-evident as I like. The illustration relies on two interesting properties of parabolas that derive from the reflection property; I'll prove them later.

Property 1. If $P$ is a point on the ("vertical") parabola, then its horizontal displacement from the vertex is the geometric mean of $4f$ and its vertical displacement from the vertex.

The illustration includes two instances of this property in the form of a classical right-triangle construction of the geometric mean.

$$\begin{align} \triangle AVC: &\quad \frac{|KV|}{|KA|} = \frac{|KC|}{|KV|} \quad\to\quad |KV|^2=|KA||KC|\quad\to\quad h^2=4f(c+k) \tag{1} \\[6pt] \triangle GSM: &\quad \frac{|VS|}{|VG|} = \frac{|VM|}{|VS|} \quad\to\quad |VS|^2=|VG||VM| \quad\to\quad s^2=4fk \tag{2} \end{align}$$

From these, we may conclude $s^2 = h^2 - 4fc$, so that the $x$-coordinates of $R_{\pm}$ ---that is, the roots of the quadratic polynomial--- have the form $$h\;\pm\;\sqrt{h^2-4fc} \tag{3}$$

(As an aside: Let the circumcircle $\bigcirc R_{+} R_{-} C$ meet the $y$-axis again at, say, $D$. Then the power of a point theorems, applied to the origin with respect to this circle, imply $$|OR_{+}||OR_{-}| = (h+s)(h-s) = c\cdot 4f = |OC||OD|$$ If we could show independently that $|OD| = 4f$, then we could reason conversely to get $(3)$ without the separate geometric means. I don't see an obvious way to make that association, however ... although little about this approach is obvious.)

Now, $(3)$ bears a bit of a resemblance to the Quadratic Formula. To get it closer, we invoke another property:

Property 2. If $P$, and distinct points $Q_{+}$ and $Q_{-}$, are on a ("vertical") parabola, such that the horizontal displacement from $P$ to each $Q$ is $f$, then the vertical displacement between the $Q$s is the distance from $P$ to the axis of the parabola.

In the figure above, $C$ plays the role of $P$, and $B_\pm$ the roles of $Q_{\pm}$. Since our parabola represents $y=ax^2+bx+c$, we have that $B_{\pm}$ is at (signed) distance $af^2\pm bf+c$ from the $x$-axis; thus, the vertical displacement between them is simply the difference of these expressions. By Property 2, we can write $$h = \left(\;af^2-bf+c\;\right) - \left(\;af^2+bf+c\;\right) = -2bf \tag{4}$$ (Recall that $b$ is non-negative here.) Therefore, $(3)$ becomes $$-2bf\;\pm\;\sqrt{4b^2f^2-4cf} \tag{5}$$ which we can write as $$2f\left(\;-b \pm \sqrt{b^2-\frac{c}{f}}\;\right) \tag{6}$$ In light of the "known" observation that $a = \dfrac{1}{4f}$ (there's that $(-1)$-dimensionality we needed!), we see $$\frac{1}{2a}\left(\;-b\pm\sqrt{b^2-4ac}\;\right) \tag{7}$$ so that we do, in fact, have the Quadratic Formula. $\square$

I'm a little disappointed in the algebraic manipulations required in this demonstration. Perhaps a second pass at the argument, drawing from some more sophisticated geometric properties of parabolas, will streamline things.

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Here are proofs of the Properties ...

Property 1.

Here, $\overline{DW}$ is the directrix of the parabola, so that $\triangle PFD$ is isosceles. The reflection property of parabolas implies that the tangent at $P$ bisects the angle at $P$; it therefore also bisects base $\overline{FD}$ at a point $M$ that, by a simple similarity argument, also serves as the midpoint of $\overline{BV}$. From similar subtriangles within $\triangle PMD$, we have $$\frac{|BM|}{|BD|}=\frac{|BP|}{|BM|} \quad\to\quad \left(\frac12 q\right)^2=fp \quad\to\quad q^2 = 4f\cdot p$$ giving the result. $\square$

Property 2.

Again, $\overline{DW}$ is the directrix. This time, we use the reflection property relative to $P$ to conclude that the tangent at $P$ is perpendicular to $\overline{FD}$. It is "known" that chord $\overline{Q_{+}Q_{-}}$ is parallel to that tangent. With a little angle chasing, we find that we may conclude $\triangle Q_{+}QQ_{-}\cong \triangle FWD$, and the property follows. $\square$

Answer 4

The coefficients $a,b,c$ of the quadratic equation $ax^2+bx^2+c=0$ are not very geometric, so let's work with some slightly different variables that do have a geometric meaning: \begin{align} \alpha &= -\frac b{2a}, & \beta &= -\frac cb, & \gamma &= c. \end{align} In reverse order, $C=(0,\gamma)$ is the $y$-intercept of the parabola, $B=(\beta,0)$ is the point where the tangent through $C$ meets the $x$-axis, and $A=(\alpha,0)$ is the point on the $x$-axis with the same $x$-coordinate as the parabola's focus. The parabola is specified via $\alpha,\beta,\gamma$, and we need to find the points $P$ and $Q$ where it crosses the $x$-axis.

^{Blue: given data, gray: constructed, green: equal quantities, red: desired roots}

Denote the focus by $F$ and the intersection of the directrix and the $y$-axis by $D$.

1. Construct the line $CF$ using the property of the parabola that the tangent $CB$ bisects $\angle OCF$. Obtain $F$ as the intersection of $CF$ and the vertical through $A$.

2. Obtain $D$ using the fact that $C$ is equidistant from $F$ and $D$. The directrix is the horizontal through $D$, and is at distance $|OD|$ from the $x$-axis.

3. Obtain $P$ and $Q$ as the points on the $x$-axis at distance $|OD|$ from $F$.

$P$ and $Q$ are equidistant from $F$ and the directrix, and so lie on the parabola.

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To derive the quadratic formula from this, we take an additional step, which may or may not be acceptable from a pure Euclidean-geometry perspective: We note that moving $C$ along the $y$-axis doesn't change the location of the roots, since it just scales the parabola vertically about the $x$-axis. Therefore, we may choose $C$ freely to simplify the construction.

In particular, let us take $C=(0,\beta)$. Then $\angle OCB=45^\circ$, so the line $CF$ is horizontal, and $F=(\alpha,\beta)$. Now $|CD|=|CF|=\alpha$, so $|OD|=\alpha-\beta$. The right triangle $\triangle AFP$ has hypotenuse $|FP|=|OD|=\alpha-\beta$ and vertical side $|AF|=\beta$, so the horizontal side is $|AP|=\sqrt{(\alpha-\beta)^2-\beta^2}=\sqrt{\alpha^2-2\alpha\beta}$; the same is true for $|AQ|$. Therefore, \begin{align} \{|OP|,|OQ|\} &= |OA| \pm |AP| \\ &= \alpha \pm \sqrt{\alpha^2-2\alpha\beta}. \end{align}

Plug in the values of $\alpha$ and $\beta$ from above, and you obtain the quadratic formula.