The methods of finding roots of a quadratic equations are quite easy and are very well understood. There are all kinds of approaches available, like algebraic, geometrical, graphical etc. The most common approach is algebraic, where we use the quadratic formula, given by

$$ax^2 + bx + c = 0 ~~~ \Longrightarrow ~~~ x = \frac{-b \pm \sqrt{b^2-4ac}}{2a} .$$

This formula gives both real and complex roots, depending on the sign of $(b^2-4ac)$. Recently, a simpler algebraic formula has been observed, which we shall discuss in a different article. Here we are interested in a graphical approach to obtain the roots of a quadratic equation.

# Real Roots

Suppose we need to find the solutions of the equation

$$x^2 - 5x + 6 = 0.$$

Using the quadratic formula given above, we can obtain the roots as

$$x = \frac{5 \pm \sqrt{5^2-4 \cdot 1 \cdot 6}}{2} ~~~ \Longrightarrow ~~~ x_1 = 2, ~~ x_2 = 3$$

Now, we do the same using the graphical method following these steps-

• First we plot this quadratic equation, which happens to be a parabola.
• We mark the points where the curve meets the horizontal axis, called $x-$axis. These points are labelled as $x_1$ and $x_2$. We can see from the graph, that their values are "2" and "3" respectively. These are the two solutions of the given quadratic equation, and they match with the result obtained by the quadratic formula.

So, this method is quite easy, and visually intuitive. Let us now move to the case when the roots are complex.

# Complex Roots

Since the previous equation did not have complex roots, we have to use another one. We can modify the previous quadratic equation to look like this

$$x^2 - 5x + 7 = 0.$$

We just changed the constant term from "6" to "7". Now let's use the quadratic formula to obtain the roots

$$x = \frac{5 \pm \sqrt{5^2-4 \cdot 1 \cdot 7}}{2} ~~~ \Longrightarrow ~~~ x_1 = \frac{5 + i\sqrt{3}}{2}, ~~ x_2 = \frac{5 - i\sqrt{3}}{2}.$$

We see that the roots are complex. Now, let us try to find these roots by using the graphical method. We again, plot the graph and get -

This graph looks a bit different from the previous case - the curve does not meet the horizontal axis. This is indicative of the fact, that there are no real roots of this equation.

This is usually, the end of story, for the graphical solutions of a quadratic equation with complex roots. But, there is one relatively obscure little idea hiding in this curve!

# Complex Roots, Revisited

Before we get back to the graph, let us do a bit of algebraic manipulation to the quadratic equation. We have obtained the complex roots, by using the quadratic formula as

$$x_1 = \frac{5 + i\sqrt{3}}{2}, ~~ x_2 = \frac{5 - i\sqrt{3}}{2}$$

If these are the solutions, then we can factor the quadratic equation as

\begin{aligned}f(x) = x^2 - 5x + 7 &= \left(x - x_1\right)\left(x - x_2\right)\\ &= \left(x - \frac{5 + i\sqrt{3}}{2}\right)\left(x - \frac{5 - i\sqrt{3}}{2} \right) \\ &= \left(x - \frac{5}{2} - \frac{i\sqrt{3}}{2}\right)\left(x - \frac{5}{2} + \frac{i\sqrt{3}}{2} \right) \\ &= \left(x - \frac{5}{2} \right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2 \end{aligned}

In this particular form, we can see that the real part of the solution is embedded in the first term, while the imaginary part is embedded in the second term. Now, we are going to see how to extract these solutions graphically.

## The Mighty Ruler

Let us rewrite the last equation here

$$f(x) = x^2 - 5x + 7= \left(x - \frac{5}{2} \right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2$$

Now, it is clear that the right hand side of this equation is a positive quantity for all values of $x$. We also observe that if we have $x=5/2$, the value of the expression is minimum. This just follows from the fact that the first term being a square of a quantity, is never negative; so its least possible value must be zero, which occurs at $x=5/2$.

This suggests that if we can find a value of $x$ where the curve has a minimum value, we have actually found the real part of the solutions!

Let us see this on the graph

We can use a standard ruler along the vertical direction passing through the point of minimum. The point where the ruler intersects the horizontal axis, is the location of the real part of the complex solution.

Quite nice!

Next, we need to locate the imaginary part. To do it, we go back to algebra for a moment. We again look at the equation

$$f(x) = \left(x - \frac{5}{2} \right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2.$$

In this equation, if we put $x=5/2$, we get

$$f\left(5/2 \right) = 0 + \left(\frac{\sqrt{3}}{2}\right)^2 \Longrightarrow \left(\frac{\sqrt{3}}{2}\right) = \sqrt{f\left(5/2 \right)}$$

We know that the values of $f(x)$ are located on the vertical axis. This means, if can find the point on the vertical axis corresponding to $x=5/2$, we have obtained the square of the imaginary part. This looks good, but we haven't got the imaginary part yet, only its square. Also, there is no obvious way to obtain square roots of numbers using just a ruler.

Let us see if we can do better. We re-write the equation as

$$f(x) = \left(x - \frac{5}{2} \right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2.$$

Let us assume that for some value of $x=x_0$, we have

$$f(x_0) = 2 \left(\frac{\sqrt{3}}{2}\right)^2.$$

This means, we can say that

$$f(x_0) = 2 \left(\frac{\sqrt{3}}{2}\right)^2 = \left(x_0 - \frac{5}{2} \right)^2 + \left(\frac{\sqrt{3}}{2}\right)^2,$$
which leads to the following equality

$$\left(\frac{\sqrt{3}}{2}\right)^2 = \left(x_0 - \frac{5}{2} \right)^2 \Longrightarrow \frac{\sqrt{3}}{2} = \pm \left(x_0 - \frac{5}{2} \right).$$

This result suggests that, if we are actually able to find this point $x_0$, we can add (or subtract) $5/2$ from it, giving us the imaginary part. In this way, we do not have to evaluate the square roots, and a ruler is just enough. Let us see how to do it on a graph.

1. Locate the point $P(5/2)$, which represents the real part of the solution as discussed earlier.
2. Place the ruler in the horizontal direction passing through the point of minimum. The point where it meets the vertical axis, marked as $Q$, is the point $f(5/2)=(\sqrt{3}/2)^2$; which means we have located the square of the imaginary part.
3. Measure the length of $OQ$ using the ruler, and mark a point $S$ such that $OS = 2 \times OQ$
4. Place the ruler in the horizontal direction passing through the point $S.$ Mark the point where the ruler meets the curve. Actually, the ruler will meet the curve at two points, and we can choose either one. Here we choose the one on the right side of the curve.
5. Place the ruler in the vertical direction passing through this point. Mark the point where the ruler meets the horizontal axis. According to the algebra before, this point happens to be $x_0.$

Here we have obtained the point $x_0$ as required. Therefore, we can finally extract the imaginary part by simply measuring the length of the line segment $PR$, which really corresponds to $x_0-5/2$. According to the algebra before, this subtraction directly gives the imaginary part of the complex roots.

That's it! We have found a graphical method to obtain the real and complex roots of any quadratic equation, using just an ordinary ruler.

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