The goal of this investigation is to explore the relationship between coefficients of the terms in quadratic equations and the roots of these equations. By thoroughly examining the ideas presented, it is the hope that students will develop a deeper understanding of quadratic equations and their solutions.
It has now become a rather standard exercise, with available technology, to construct graphs to consider the equation
ax2 + bx + c = 0
and to overlay several graphs of
for different values of a, b, or c, holding the other two constant. From these graphs students can discuss how the patterns for the roots of
change with the coefficients of the terms. For example, if we set
for b = -3, -2, -1, 0, 1, 2, 3, and overlay the graphs, the following picture is obtained.
We can discuss the "movement" of a parabola as b is changed. The parabola always passes through the same point on the y-axis ( the point (0,1) with this equation). For b < -2 the parabola will intersect the x-axis in two points with positive x values (i.e. the original equation will have two real roots, both positive). For b = -2, the parabola is tangent to the x-axis and so the original equation has one real and positive root at the point of tangency. For -2 < b < 2, the parabola does not intersect the x-axis -- the original equation has no real roots. Similarly for b = 2 the parabola is tangent to the x-axis (one real negative root) and for b > 2, the parabola intersects the x-axis twice to show two negative real roots for each b.
So the value of b, the coefficient to the x term, determines the parabola’s location along the x axis, as well as the number and value of the roots of the equation.
Consider the locus of the vertices of the set of parabolas graphed from
Click here to see how the parabola y = (-x)2 + 1 is formed from the locus of the vertices. Is there a general statement we can make about the relationship between y = x2 + bx + 1 and y = (-x)2 + 1 ?
Now let’s look at what happens if we vary the constant, c.
X2 + X + 1 x2 + x + 0 x2 +x - 1 x2 +x – 2 x2 + x + 2
Varying the constant term moves the parabola up and down the y axis. In doing so, it also plays a role in the determination of the number of real roots. When c is negative, as in x2 + x -1, the graph has two real roots (one positive and one negative). However, when c is positive, there are no real roots.
And if we vary a, the coefficient to the x2 term, we’ll see that the value of a determines the width of the parabola, and consequently the value of the roots. While always having two real roots, notice how their values change.
X2 + x + 0 3x2 + x + 0 5x2 + x + 0 0.5x2 + x + 0
Graphs in the xb plane.
As the investigation continues, we explore quadratic equations from a whole new perspective.
Consider the quadratic equation x2 + bx + 1 = 0. This is its graph in the XB plane.
But how do we come up with this graph? Why is it shaped like a rational function? It certainly doesn’t look like a quadratic equation. How can we make sense of it?
Isolating the b value gives it to us. If ordinarily we find y as a function of x, in the xb plane we find b as a function of x. Solve our original quadratic equation for b:
Start Y = 0
0 = x2 + bx + 1
bx = -x2 -1
b = -x2 - 1/x
b = -x – (1/x)
And now we call b the name y, just so that we are operating on familiar terms again. If we graph y = -x – (1/x) we do, in fact, get the graph pictured above. In the familiar Cartesian plane we graph a function by isolating and solving for y. It makes sense that we isolate and solve for b in the xb plane.
If we overlay the line (for instance, b=3) on the graph above, the intersection points with the graph in the xb plane (if they exist) correspond to the roots of the original equation (x2 + bx + 1) for that value of b.
This can be understood in the following way: The roots of the original quadratic equation when b=3 (x2 + 3x + 1), are -.382 and -2.618. Taking the function in the xb plane, (y = -x – (1/x)), replacing the x with the root values, and setting it equal to our value of b, we have
b = - x – (1/x) and b = - x – (1/x)
3 = - (-.382) – (1/-.382) 3 = - (-2.618) – (1/-2.618)
3 = .382 + 1/.382 3 = 2.618 + 1/2.618
3 = .382 + 2.618 3 = 2.618 + .382
3 = 3 3 = 3
which is, of course, true. So the line representing the b value in the xy plane becomes the x axis in the xb plane. Just as the quadratic function in the xy plane crosses the x axis at the roots -.382 and -2.618, so, too, does the function in the xb plane cross the horizontal axis (b) at the same root values of -.382 and -2.618.
X2 + 3x + 1 -X – (1/x) Y = 3
For each value of b we select, we get a horizontal line. It is clear on a single graph that we get two negative real roots of the original equation when b > 2, one negative real root when b = 2, no real roots for -2 < b < 2, one positive real root when b = -2, and two positive real roots when b < -2.
This is precisely what we found when examining the original quadratic equation, x2 + bx + 1.
Graphs in the xc plane
In the following example the equation
is considered. If the equation is graphed in the xy plane, it is easy to see that the curve will be a parabola. For each value of c considered, its graph will be a line crossing the parabola in 0, 1, or 2 points -- the intersections being at the roots of the original equation for that value of c.
x2 + 5x + 1
At c = 1, for instance, the equation x2 +5x + 1 has two negative roots – approximately -0.2 and -4.8. At c = -6.25, the equation has just one real root. At values of c less than -6.25, there are no real roots.
The factors involved in the study of quadratic equations are often taught separately, with few connections made to their interrelatedness. By using these examples to explore quadratic relationships more fully, it is the hope that students will develop a deeper understanding of them.