The roots of a Quadratic Equation

By: Diana Brown

It has now become a rather standard exercise, with available technology, to construct graphs to consider the equation

and to overlay several graphs of

for different values of a, b, or c as the other two are held constant. From these graphs I am going to explore the roots of

First lets set a and c equal to 1 and let b = -4, -3, -2, -1, 0, 1, 2, 3, 4.  See the below graph:

Notice in the graph above 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.

Remember the roots (sometimes also called "zeros") of a quadratic equation, f(x) =0, are the values of x for which the equation is satisfied.  Lets derive a formula to find the solution of a quadratic equation:

Consider the equation

The roots x can be found by completing the square,

Solving for x then gives

Note that this is the quadratic formula and this formula is used to find the roots of a quadratic equation.

Now let’s look at the locus of the vertices of the set of parabolas graphed from above:

 EQUATION VERTEX y = x² - 4x + 1 (2, -3) y = x² - 3x + 1 (1.5, -1.25) y = x² - 2x + 1 (1, 0) y = x² - x + 1 (0.5, 0.75) y = x² + 1 (0, 1) y = x² + x + 1 (-0.5, 0.75) y = x² + 2x + 1 (-1, 0) y = x² + 3x + 1 (-1.5, -1.25) y = x² + 4x + 1 (-2, -3)

See the graph of the locus of the vertices of the parabolas:

Notice that the locus is a parabola.  We can see that the vertex of the above parabola is at (0, 1) and that each point is symmetric to each other about the line x=0.  Also the x intercepts are at the points (1, 0) and (-1, 0).  If we set y = to the roots of the equation we obtain:

y= (x – 1) (x + 1)

y = x² - 1

and, since the parabola is opening down therefore a must be negative so lets see what happens when we graph the equation y =  -x² + 1 on the above graph:

As you can see in the above graph, the locus of the vertices of the set of parabolas graphed from y = x² + bx + 1, in the beginning of the exploration is the parabola:

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