 | ON-Math Spring 2003 | Volume 1, Number 3 | |
Audrey WeeksWe are very familiar with relating the real root(s) of a quadratic
equation to the x-intercept(s) of the graph
of its related function, a parabola. But what about the quadratic whose
roots have an imaginary part and whose parabola does not
intersect the x-axis? Did you know that
we still can make a graphical connection?
As an overview, here is what follows:
- A detailed graphical connection to a parabola’s complex roots
using an interactive applet
- An explanation of the complex number plane along
with some of the history of its development
- A proof of why this connection between the graph
of a parabola and its complex roots holds
- A student worksheet (with solutions)for
additional exploration in algebra classes
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When first learning to solve a quadratic equation of the form ax2
+ bx + c = 0 (a ≠ 0), we generally
have our students explore four possible techniques. Three of these techniques
are algebraic — solving by factoring, solving by completing the square,
and solving by the quadratic formula (this is still completing the square,
but we jump to the final line saving ourselves many steps). However, before
using these abstract solution techniques, it is advised that we first
give our students a visual image of the situation, so we approach the
problem graphically. We consider the graph of the related function y
= ax2 + bx + c. This graph is, of course, a parabola.
Here the value of the trinomial (the y values)
are allowed to run the gamut beginning with the height of the parabola’s
vertex and continuing unbounded in either a positive direction (if a
> 0) or a negative direction (if a <
0). The only coordinates that solve our original equation, however, are
the ones for which the trinomial equals 0 (i.e., y
= 0). Hence, the x-intercepts of the parabola (the graph of the related
function) are the solutions to our original quadratic equation.
Figure 1 |
As an example, to find the solutions to the quadratic
equation 0 = x2 – 2x
– 3 by a graphical approach, we graph the related function y
= x2 – 2x
– 3 (See figure 1). Each ordered pair on this graph is
a solution of the y = x2
– 2x – 3 function but only
ordered pairs of the form (k, 0) is
a solution of the original 0 = x2
– 2x – 3 equation. These points
are the x-intercepts of the graph, (–1, 0) and (3, 0). Therefore,
x = –1 or x
= 3 are solutions of our original equation. |
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Solving quadratic equations by graphing is such a clear visual
representation of the process that our students may wonder, especially
if the repetition of solving or and plotting points is alleviated
by a graphing calculator, why learn another approach? We may caution,
as in figure 2, that the x-intercepts
of the related function may not fall on convenient integers. This
function expression is y = x2
– 3x + 1. It would be highly
unlikely, to say the least, to estimate that these x-intercepts
are at  .
Of course, our students may counter with the suggestion that the
decimal approximations for these roots can be ascertained, to a
high degree of precision, by using the graphing calculator .
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Figure 2 |
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Figure 3
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Rather than quibble over decimal places, we execute the knockout
punch. Consider a parabola like the one in figure 3. Its
function expression is y = (x
– 3)2 + 4 or y = x2
– 6x + 13. It has no x-intercepts,
so we may suggest that we must learn an algebraic approach.
We explain that this function is related to a quadratic equation
having no real roots. Its solutions are in the set of complex
numbers having an imaginary component and, therefore, cannot be
found on a graph. Employing the quadratic formula, we find its roots
to be x = (3 + 2i)
or x = (3 – 2i),
where i =  .
Is this situation really the breakdown of the graphical approach?
No, indeed it is not!
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