### A New
Look at Quadratics

### by

### Mindy
Swain

**Assignment
3**
**Investigation
#2: Graphs in the xb plane**
Consider the equation
, where a = 1 and c = 1.

Now graph this relation
in the xb plane. So, . We have the following
graph.

If we take any particular
value of b, say b = 4, and overlay this equation on the graph
we add a line parallel to the x-axis. If it intersects the curve
in the xb plane the intersection points correspond to the roots
of the original equation for that value of b. We have the following
graph.

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.

Consider the case when
a=1 and c = - 1. So, .

Now graph this relation
in the xb plane. Now . We have the following
graph.

If we take any particular
value of b, say b = 4, and overlay this equation on the graph
we again add a line parallel to the x-axis. We have the following
graph.

For each value of b
we select, we get a horizontal line. It is clear on a single graph
that we get one negative real root and one positive real root
for all values of b.

Now let's consider other
values of c and generalize. Click play below to see an animation
for values of c. The blue graph is for c = 1, the purple graph
is for c = - 1 and the red is for varying values of c.

From this animation
we can conclude that different c-values have a different number
of solutions for b.

If c < 0, there will
always be one positive real root and one negative real root for
b.

If c = 0, there will
be exactly one real root for b.

If c > 0, there are
some vales of b with two real roots and some values of b with
no real roots.

Let's look at this
closer. Considering the discriminant of the quadratic formula
when a = 1 and c > 0, we have . When
the discriminant is negative is when there is no real solution,
so when there is no real solution for
the values of b such that . We
can conclude for the original equation when...

:
two negative real roots;

:
one negative real root;

:
no real roots;

:
one positive real root;

:
two positive real roots.

**Return**