I read that if a parabola has no real roots, then its complex roots can be visualized by graphing the same parabola ($ax^2 + bx + c$) with $-a$ and then finding the roots of that, then using those roots as the diameter of a circle. Then, the "top" and "bottom" of the circle will be the complex roots of the parabola (in the complex plane).

I don't know if I explained that in [enough] detail, but I'm trying to prove this. I've been trying to generalize an equation for all the parabolas with no real roots, and I seem to be getting somewhere, but it seems like I'm over-complicating this.

How would you approach this? Any suggestions would be appreciated.

A proof of this construction would also be an acceptable answer. Thank you.


There is a statement along these lines, but it doesn't work to simply negate $a$ when the quadratic is in standard form. You have to start by completing the square: $$ ax^2 + bx + c \;=\; a(x-h)^2 + k. $$ Then the roots of the quadratic $$ -a(x-h)^2 + k $$ will have the properties you describe. In particular, if we write the roots of this quadratic as $h + r$ and $h - r$, then the roots of the original quadratic are $h + ri$ and $h - ri$.

By the way, observe that the auxiliary quadratic described above has the same vertex as the original quadratic, but its graph has been reflected across the horizontal line that goes through this point. This gives a nice geometric way of "picturing" the complex roots of a quadratic by looking at the graph.


Either I'm misunderstanding what you wrote, I did a mistake below, or the construction does not work.

Assume that $a\gt 0$ (we may as well, since the roots of $ax^2+bx+c$ are the same as the roots of $-ax^2-bx^2-c$).

The roots of $-ax^2+bx+c$ are the same as the roots of $ax^2-bx-c$, and are given by $$\frac{b+\sqrt{b^2+4ac}}{2a}\quad\text{and}\quad\frac{b-\sqrt{b^2+4ac}}{2a}.$$ If you are assuming that $ax^2+bx+c$ has no real roots (so that $b^2-4ac\lt 0$), then this has real roots, since $b^2-4ac\lt 0$ implies that $ac\gt 0$, so $b^2+4ac\gt 0$.

The circle whose diameter goes from the point $\left(\frac{b-\sqrt{b^2+4ac}}{2a},0\right)$ to $\left(\frac{b+\sqrt{b^2+4ac}}{2a}\right)$ has center at $\left(\frac{b}{2a},0\right)$, and radius $\frac{\sqrt{b^2+4ac}}{2a}$. So the coordinates of the "top" and "bottom" of that circle are $$\left(\frac{b}{2a},\frac{\sqrt{b^2+4ac}}{2a}\right)\quad\text{and}\quad\left(\frac{b}{2a},-\frac{\sqrt{b^2+4ac}}{2a}\right).$$ That is, they correspond to the complex numbers $$\frac{b}{2a} + \frac{\sqrt{b^2+4ac}}{2a}i\qquad\text{and}\qquad \frac{b}{2a}-\frac{\sqrt{b^2+4ac}}{2a}i.$$ These are the roots of $ax^2+bx+c=0$ if and only if they are the roots of $x^2+\frac{b}{a}x + \frac{c}{a}=0$, if and only if they add up to $-\frac{b}{a}$ and multiply to $\frac{c}{a}$. But they add up to $\frac{b}{a}$, not $-\frac{b}{a}$; and their product is $$\left(\frac{b}{2a} + \frac{\sqrt{b^2+4ac}}{2a}i\right)\left(\frac{b}{2a}-\frac{\sqrt{b^2+4ac}}{2a}i\right) = \frac{b^2}{4a^2} + \frac{b^2+4ac}{4a^2} = \frac{b^2+2ac}{2a^2}\neq \frac{c}{a}.$$

For an explicitly example, take $x^2+x+1$, which has no real roots; the complex roots are $\frac{1}{2}+\frac{\sqrt{3}}{2}i$ and $\frac{1}{2}-\frac{\sqrt{3}}{2}i$. The construction you describe begins by considering instead $-x^2+x+1$, whose roots are $\frac{1}{2}+\frac{\sqrt{5}}{2}$ and $\frac{1}{2}-\frac{\sqrt{5}}{2}$. The circle whose diameter goes from $(\frac{1}{2}-\frac{\sqrt{5}}{2},0)$ to $(\frac{1}{2}+\frac{\sqrt{5}}{2},0)$ has center at $(\frac{1}{2},0)$ and radius $\frac{\sqrt{5}}{2}$, so the "top" and "bottom" of the circle will be $(\frac{1}{2},\frac{\sqrt{5}}{2})$ and $(\frac{1}{2},-\frac{\sqrt{5}}{2})$, which are not the complex roots of $x^2+x+1$.

The construction works if $b=0$: in that case, instead of graphing $ax^2+c$, with $\frac{c}{a}\gt 0$, we graph $-ax^2+c$; the zeros are $\pm\sqrt{\frac{c}{a}}$, so the circle in question will be centered at the origin and have radius $\sqrt{\frac{c}{a}}$, so the complex numbers corresponding to the "top" and "bottom" are exactly $i\sqrt{\frac{c}{a}}$ and $-i\sqrt{\frac{c}{a}}$, which are the roots of $ax^2+c$ when $\frac{c}{a}\gt 0$.

So this suggests that for the general case you should first complete the square and then change the sign of the entire squared factor. For my example, beginning with $x^2+x+1$, first complete the square: $$x^2 + x + 1 = \left(x^2 + x + \frac{1}{4}\right) + \frac{3}{4} = \left(x + \frac{1}{2}\right)^2 + \frac{3}{4}.$$ Then graph $-(x+\frac{1}{2})^2 + \frac{3}{4} = -x^2 -x + \frac{1}{2}$ and proceed as you describe. This will work: it amounts to first doing a horizontal shift so that we are dealing with a quadratic of the form $Az^2 + C$ (where $z=x+k$ for some $k$), and then proceeding as above. But you'll note that the resulting equation is not obtained simply by changing the sign of $a$: the values of $b$ and $c$ are also changed.

You can check that this modified idea works by doing essentially what I did above, but showing it does work.


We give a construction of the complex roots, focusing on the geometry.

The general quadratic equation has the shape $$ax^2+bx+c=0,$$ where $a \ne 0$. But if we are only interested in the geometry of the roots, it does no harm to divide every coefficient by $a$, obtaining the simplified form $$x^2+px+q=0.$$
Geometrically, this corresponds to a scaling in the $y$-direction.

Consider the parabola with equation $y=x^2+px+q$. Move the parabola sideways so that the vertex of the parabola lies on the $y$-axis. Then the equation of the shifted parabola assumes the simple shape $$y=x^2+d^2,$$ where $d>0$. We have written the constant term of the shifted parabola as $d^2$, since because the parabola does not cross the $x$-axis, the constant term must be positive.

The vertex of the parabola is at $(0,d^2)$. Draw a horizontal line at distance $d^2$ above the vertex. This line has equation $y=2d^2$, so it meets the parabola at the points with $x$-coordinates $x=\pm d$.

Draw a circle with center halfway between these two points, and passing through them. So the circle has radius $d$. Lower this circle to make a new circle $C$ with center the origin.

The "top" and "bottom" of this new circle $C$ are at $(0,d)$ and $(0,-d)$. If we use complex numbers to represent them, they are $0+di$ and $0-di$, the complex roots of $x^2+d^2=0$.

Now shift the parabola back to its original position, dragging everything we have constructed along. Then the roots of $x^2+d^2=0$ are dragged along to become the roots of the original equation $x^2+px+q=0$.

The shifting was done to make the algebra simple. But now we can give a geometric recipe for constructing the complex roots of a quadratic equation that has no real roots, say with positive coefficient for $x^2$.

(i) Draw the appropriate parabola.

(ii) Draw a line which is just as far above the vertex of the parabola as that vertex is above the $x$-axis. Suppose that this line meets the parabola at points $P$ and $Q$.

(iii) Draw the circle with center midway between $P$ and $Q$ and passing through $P$ and $Q$.

(iv) Lower this circle so that it becomes a circle $C$ with center on the $x$-axis.

(iv) The complex roots of our equation are at the "top" and "bottom" of circle $C$.


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