Explore
Quadratic Equation and Its Roots

By
Graphing Calculator

Najia
Bao

When technological tools are
available, students can focus on decision making, reflection, reasoning, and
problem solving (NCTM, 2000).

This lesson is an excellent
opportunity for using technology to explore quadratic equations and roots.

**Investigation 1-3**

** **

For the general quadratic equation y
= ax^{2} + bx + c, if we set a = 1 and c =1, that is to say, y = x^{2}
+ bx + 1, and then we set b = -3, -2, -1, 0, 1, 2, 3, we can see the ÒmovementÓ
of a parabola as b is changed (see fig. 1).

In figure 1, the blue curve
represents the parabola with b = 3, the red one represents the parabola with b
= 2, the purple represents the parabola with b = 1, the black represents the
parabola with b = 0. And the green, light blue, and yellow represents the
parabola with b = -1, -2, -3 respectively. It seems that when b is negative,
the parabola ÒmovesÓ to the right and when b is positive, it ÒmovesÓ to the
left. We also noticed that when b = 3, the equation has two negative real
roots. That is because the parabola intersects the x-axis at two points.
Similarly, when b = - 3, the equation has two positive real roots. When b = 2
or –2, the parabola is tangent to the x-axis, so the equation has just
one real root, which is -1 or 1 respectively. When –2 < b < 2, the equation has no real roots because the parabola
does not intersect the x-axis.

From figure 1, we also found
something else interesting. All of the parabolas passes through the point
(0,1). That is because whenever x =0, y = x^{2} + bx + 1 = 0 + 0 +1
=1.

Furthermore, we
found an amazing and interesting phenomenon that the locus of the vertices of
the set of parabolas graphed from y = x^{2} + bx + 1 is the parabola y
= -x^{2} + 1 (see the red parabola in fig. 2).

Figure 2

Now letÕs substitute y for b. So x^{2} + bx + 1 = 0 is changed to x^{2}
+ yx + 1 = 0. Then we
get the following graph (see fig. 3).

Figure 3

Then
we add another equation y = 3, which is a line paralleling to x-axis (see fig.
4). We find that the line y =3 intersects the original curve at two points.
Therefore, the system of equations x^{2} + yx + 1 = 0 and y = 3 has two
negative real roots. From figure 4, we also find that the system of equations
has always two real roots as long as y > 2 or y < -2, and the system of equations has just a distinct
real root when y = 2 or –2. Of course, we can also imagine that when -2 < y <
2, it has no real root.

Figure 4

Next, letÕs change the value of c from 1 to –1, that
is, the original equation is changed to x^{2}
+ yx - 1 = 0 (see fig. 5). And then we add a new linear equation 2x + y = 0,
that is, y = -2x (see fig. 5). From figure 5, we find that it intersects the
curves representing the quadratic equation x^{2} + yx + 1 = 0 at two
points, thus the system of equations comprised of them has two real roots (one
positive and the other negative). Also, we find that the line 2x + y =0 does
not intersect the curves representing the quadratic equation x^{2} + yx
- 1 = 0, thus there is no real roots.

Figure 5