James W. Wilson and Andrew Tyminski

University of Georgia

In countless algebra 1 classrooms all across America, teachers show their students the standard form of a quadratic equation and then explain various ways of finding its roots. However, one thing that doesn't always happen in these classrooms is an explanation of the relationship between the standard form of the quadratic equation and its roots. That is, how do the coefficients a,b, and c effect the roots of a given equation? Can you tell just by looking at an equation whether it will have real roots, or imaginary roots? Can you tell how many roots there will be, and if so, will they be positive, or negative?

If you answered yes to any of the preceding questions, then you probably teach your students about the discriminant and how it can be used to determine the number and type of roots of a given quadratic. Good for you. What follows here is a way to show the relationship of the discriminant graphically using some simple algebra and a graphing calculator program such as Graphing Calculator. If you answered no to the above, allow me to drop some knowledge on you.

First of all, lets review exactly what the discriminant is. In the quadratic formula, the discriminant is the part under the radical sign. Namely, . The standard algebra 1 text will state that when the discriminant is positive, then there will be two real roots of the equation. If the discriminant is zero, there will be one real root, and if the discriminant is negative, this indicates no real roots. This is all well and good, but I believe our teaching would be more powerful if we could show this relationship visually in a way that makes the concept behind the "rule" make sense.

To accomplish our goal, we are going to have to look at the graphs of quadratic equations in a different way. Instead of graphing in our normal xy plane, we are going to look for relationships between a, b, and c in the xb plane. That is, the horizontal axis will represent the x variable, and the y axis will represent the variable b in . This means we will really be working with equations of the form . What we would like to do is examine each of the cases described in the text from an algebraic standpoint, and then show these same relationships graphically. So, let's begin...

If > 0, then we know > . Taking the square root of both sides we find that b > +/-2. But. because this is a compound inequality, we need to split it into two inequalities. These will be b > 2, or b < -2. So, in order for an equation to have real roots, a, b, and c will have to satisfy these equations. Let's look at some graphs of quadratic equations in the xb plane that fit these inequalities and see what is really happening.

Case 1 - Both a and c are positive

Below is a graph of the equation (remember we are in the xb plane so we can look at how a and c relate to b.)

There will be real solutions to this equation whenever any horizontal line, representing the value of the b variable, intersects the graph. For example, the line b= 5 will intersect the graph in 2 points. These points of intersection are the two roots of the equation for b =5.

We can see that for any value of b > 4, we will get two real solutions for x and they will both be negative. In the above example, when b = 5, the two roots are x = -4 and x = -1.

You may also notice that we will also get two solutions when our horizontal line intersects the bottom half of the hyperbola. That is, when x < -4. For example, using the line b = -6, we see below that it also intersects in two places indicating two real roots. In this case, both roots are positive. (approximately, x = .76392 and x = 5.23607)

Now, let's go back to our original equation in standard form and check our "rule". Remember that the rule says we should get two real roots when For , a = 1 and c = 4. Substituting into our compound inequalities implies that b > 2 or b < -2 , which means b >4 or b < -4. Remarkably, or maybe not so, these were the same parameters we decided on based upon our graphs. But, in addition to this information, the graphs were also able to tell us that when b >4, both of the solutions would be negative , and when b < -4, both solutions will be positive. Nice!

The second "rule" states that when the discriminant is zero, then there will be one real root. Algebraically, this means that . We can explore the second "rule" using the same graphs as above. From the graph below we can find a horizontal line that is tangent to the upper half of the hyperbola at its vertex, as well as one that is tangent to the bottom half at its vertex.

These two lines are the lines b = 4 and b = -4. They are tangent to, each at the vertex of the curve. When b = 4, we get one real root that is negative, namely x = -2. When b = -4, then we get one positive root, that is, x = 2. To verify our finding algebraically,using the "rule", we substitute our values for a and c into .In doing this, and solving for b, we see that either b =4 or b = -4. Once again, the graph is not only giving us a visual of this, but also helping determine the sign of the solution.

The third "rule" refers to no real solutions when < 0. Algebraically, we can show, using the same argument as above, that this implies that there will be no real solutions when b < 2 and when b > -2. Looking at the graph of , we can see that the horizontal lines for b, where - 4 < b < 4, will never intersect either of the hyperbolas. For example, the lines b = 3 and b = -2, shown below, imply that there are no real solutions to our equation for either of those values of b.

Verifying the rule algebraically we can substitute and find that there are no real roots for the equationwhen b <2, and when b >-2. Simplifying these expressions, we see the rule indicates no real solutions for -4 < b <4.

Case 2 - Both a and c are negative

Now we will look at an example where both a and c are negative. Let's look at the equation , and its graph.

Once again, in the xb plane, we will have a solution any time a horizontal line intersects either half of the hyperbola. So for any values of b > 4 or b < -4, we should get two solutions. So if , for example, b = 7, or if b = -9, we would get two points of intersection that would symbolize the roots for that given equation.

In the case of b = 7, we see that there are two positive values for x, approximately x = .313, or x = 3.19. However, when b = -9, we get two negative values for solutions. This is exactly the opposite of what we found when both a and c were positive. This makes sense of course, since that if a and c have the same sign, then -4ac will always have the same sign. This implies the solutions will either both be positive or both negative. Changing the signs of a and c simply reflected the graph across the axis, which in turn changed when the solutions will be positive or negative.

From the "rule", we should see that we get two real roots when b > 2 and when b < -2. In this particular case, substitution shows us that b >4 or b < -4, exactly what the graph shows.

We could repeat the above arguments to show the graph and the "rule" agree for one solution, or no solutions as well. In this case, one solution when b = 4 or b = -4, no solutions for -4 < b < 4.

Case 3 - a and c have different signs

This case is a bit different that the first two. Let's look at two different equations, and .

You should notice that the
graphs in the xb plane are still hyperbola, but these are much
more shallow than the other graphs we have seen. This shallowness
will have a big effect on our solutions. If we draw any horizontal
line in either of these graphs, we will see that it will __always__
have two intersections, and therefore two solutions. In both cases, there will be one solution that is
positive and one solution that is negative. If we think about
this algebraically using our "rule" we can see that
this conclusion makes sense.

If a and c have different signs, then the value of -4ac will always be positive. So we will get two solutions if > 0. If this is true, then we know > -. Since b squared is always positive, we know that this inequality will be true for any value of b.

So, we have shown the graphical representations
of the discriminant as a function of number of solutions. This
will help students to build a better understanding of the mathematics
behind solving quadratics. This will make them **leaders**
of the classroom, not just