Visual Detection of Imaginary Roots in a Parabola

A Parabola function is either presented in its standard form, y= a(x - b) + h, or in its polynomial form, y = ax + bx + c.
File:Nassim1Rev.JPG Figure 1: Parabolas (A) and (B) have real and imaginary roots, respectively.
As shown in figure 1, if a parabola intersects x-axis at x<sub>1</sub> and x<sub>2</sub>, its real roots can be read as -2 and -8 fairly accurately by visual inspection depending on the quality of its plot. However if the parabola is entirely above or entirely below the x-axis, like parabola (B), it has imaginary roots.

To predict if a parabola is expected to have real or imaginary roots one can use the discriminant (b -4ac) test by plugging the values of a, b, c coefficients from its equation, y=ax +bx+c:

1. If b - 4ac = 0 then the parabola has one real root.

2. If b - 4ac > 0 then the parabola has two real roots.

3. If b - 4ac < 0 then the parabola has two imaginary roots.


However, there is a simpler method of doing this if the standard equation of the parabola, y = a(x - b) + h, is available:

1. If h = 0 then the parabola has one real root.

2. If (a)(h) < 0 then the parabola has two real roots.

3. If (a)(h) > 0 then the parabola has two imaginary roots

The main goal in this article, nonetheless, is to show how one can read the imaginary roots off of a parabola plot on a graph paper.
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Below, the steps to read the imaginary roots of a parabola by visual inspection are given. Later, the validity of the results are verified by plugging them into its equation; and also, they are compared to the results obtained using the quadratic equation.
File:Nassim2Rev3.JPG Figure 2: The geometric interpretation of imaginary roots in the Cartesian coordinate system.
Using the example shown in figure 2, the steps to read the imaginary roots of a parabola visually are:
1. Draw line y 2h (y 4) and notice the points where it intersects the parabola.
2. Mentally, find the value of P (half distance between the points of intersection).
3. Call out the imaginary roots by plugging the values of b (vertex) and P in b ± P. (For the parabola in this figure, it will be 4 ± 2i)
At this point the original stated goal has been achieved. After this method of visual detection tried a couple of times, the values of b and P will stand out in the case a different parabolas without much additional effort.
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The following three tests are presented to verify the accuracy of 4 ± 2i detected visually.


;Test 1: Plug 4 + 2i in the parabola equation:

Y = 0.5(x - 4) +2
: = 0.5[(4 + 2i) - 4) + 2
: = 0.5(4 + 2i - 4) + 2
: = 0.5(2i) +2
: = 0.5(4i ) + 2
: = 0.5(4(-1)) + 2
: = -2 + 2
: = 0

with Y = 0, test 1 is successful and confirms the validity of the discussion
;Test 2: Plug 4 - 2i in the parabola equation:

Y = 0.5(x - 4) +2
: = 0.5[(4 - 2i) - 4) + 2
: = 0.5(4 - 2i - 4) + 2
: = 0.5(-2i) +2
: = 0.5(4i ) + 2
: = 0.5(4(-1)) + 2
: = -2 + 2
:= 0

likewise with Y = 0, test 2 is successful as well
;Test 3: Determine the roots of y = 0.5(x - 4) +2 using quadratic equation, x<sub>1</sub>, x<sub>2</sub> = (-b ± (b - 4ac) )/2a:

The equation in polynomial form is y = 0.5x - 4x + 10 with a 0.5, b - 4, and c = 10.

The test of its discriminant (the portion under the radical) yields:

b - 4ac (- 4) - 4(0.5)(10) - 4 < 0

Here the negative value confirms the fact that this parabola has imaginary roots. Proceeding to find the imaginary roots using quadratic equation:

x<sub>1</sub>, x<sub>2</sub> = (-b ± (b - 4ac) )/2a

Substituting -4 for b - 4ac, -4 for b, and 0.5 for a, we have

: = (-(-4) ± (- 4) )/2(0.5)

: = (4 ± (4i ) )/1

: = 4 ± 2i

This result is consistent with what we got from the visual inspection earlier.
The tests above establish the fact that the visual detection method discussed here is valid. In later updates, the practical application of this finding will be discussed.

In figure 2, the empty circles on the x-axis are the physical locations of the imaginary roots for the parabola on display. These imaginary roots are displayed via empty circles intentionally to distinguish them from typical real roots. These empty circles are read as:

x<sub>1</sub> = 4-2i

x<sub>2</sub> = 4+2i

They are not read like real roots as x<sub>1</sub> 2 and x<sub>2</sub> 6

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The diagrams in this article plot the points indicating the imaginary roots using small empty circular dots. Real intersections are indicated by circular filled in dots. The diagrams are plotted in the realCartesian coordinate system, and no imaginary axis is shown. This method is an alternate way of plotting imaginary roots which can also be plotted in i-x coordinates.
 
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