# The Quadratic Formula

##
**Intercepts**

In **Example 13,** we used the quadratic formula to
find the solutions of x^{2} − 2x − 2 =

0. These solutions, and their approximations, are shown in **equation (14).**
It is

important to make the connection that the solutions in **equation (14) **are
the zeros

of the quadratic function g(x) = x^{2} −2x−2. The zeros also provide the
x-coordinates

of the x-intercepts of the graph of g (a parabola). To emphasize this point,
let’ s draw

the graph of the parabola having the equation g(x) = x^{2} − 2x − 2.

First, complete the square to place the quadratic function in vertex form. Take
half

the middle coefficient and square , as in [(1/2)(−2)]^{2} = 1; then add
and subtract this

term so the equation remains balanced.

g(x) = x^{2} − 2x − 2

g(x) = x^{2} − 2x + 1 − 1 − 2

Factor the perfect square trinomial , then combine the
constants at the end.

g(x) = (x − 1)^{2} − 3

This is a parabola that opens upward. It is shifted to the right 1 unit and down
3 units.

This makes it easy to identify the vertex and draw the axis of symmetry, as
shown in

**Figure 1**(a).

It will now be apparent why we used our calculator to approximate the solutions

in (**14**). These are the x-coordinates of the x-intercepts. One x-intercept
is located at

approximately (−0.73, 0), the other at approximately (2.73, 0). These
approximations

are used to plot the location of the intercepts as shown in** Figure 1**(b).
However, the

actual values of the intercepts are ((2−)/2,
0) and ((2+)/2, 0), and these exact

values should be used to annotate the intercepts, as shown in **Figure 1(**b).

Finally, to find the y-intercept, let x = 0 in g(x) = x^{2} − 2x − 2.
Thus, g(0) = −2

and the y-intercept is (0,−2). The y-intercept and its mirror image across the
axis of

symmetry are both plotted in **Figure 1**(c), where the final graph of the
parabola is

also shown.

(a) Plotting the vertex and axis of symmetry. |
(b) Adding the x-intercepts provides added accuracy. |
(c) Adding the y-intercept and its mirror image provides an excellent final graph. |

We’ve made an important point and we pause to provide emphasis.

Zeros and Intercepts. Whenever you use the quadratic formula to solve the

quadratic equation

ax

^{2}+ bx + c = 0,

the solutions

are the zeros of the quadratic function

f(x) = ax^{2} + bx + c.

The solutions also provide the x-coordinates of the x-intercepts of the graph of f.

We need to discuss one final concept.

**The Discriminant
**

Consider again the quadratic equation ax

^{2}+ bx + c = 0 and the solutions (zeros)

provided by the quadratic formula

The expression under the radical , b

^{2}−4ac, is called the discriminant, which we denote

by the letter D. That is, the formula for the discriminant is given by

D = b

^{2}− 4ac.

The discriminant is used to determine the nature and number of solutions to the quadratic

equation ax

^{2}+bx+c = 0. This is done without actually calculating the solutions.

Let’s look at three key examples.

Example 15. Consider the quadratic equation

x

^{2}− 4x − 4 = 0.

Calculate the discriminant and use it to determine the nature and number of the

solutions.

Compare x

^{2}− 4x − 4 = 0 with ax

^{2}+ bx + c = 0 and note that a = 1, b = −4, and

c = −4. The discriminant is given by the calculation

D = b

^{2}− 4ac = (−4)

^{2}− 4(1)(−4) = 32.

Note that the discriminant D is positive; i.e., D > 0.

Consider the quadratic function f(x) = x^{2} −4x−4, which can be written
in vertex

form

f(x) = (x − 2)^{2} − 8.

This is a parabola that opens upward. It is shifted to the right 2 units, then
downward

8 units. Therefore, it will cross the x-axis in two locations. Hence, one would
expect

that the quadratic formula would provide two real solutions (x-intercepts).
Indeed,

Note that the discriminant, D = 32 as calculated above, is
the number under the square

root. These solutions have approximations

and

which aid in plotting an accurate graph of f(x) = (x − 2)^{2}
− 8, as shown in **
Figure 2.**

Figure 2. If the discriminant is positive,

there are two real x-intercepts.

Thus, if the discriminant is positive, the parabola will have two real
x-intercepts.

Next, let’s look at an example where the discriminant equals zero.

Example 16. Consider again the quadratic
equation ax^{2} + bx + c = 0 and the

solutions (zeros) provided by the quadratic formula

The expression under the radical, b^{2} −4ac, is
called the discriminant, which we denote

by the letter D. That is, the formula for the discriminant is given by

D = b^{2} − 4ac.

The discriminant is used to determine the nature and number of solutions to the
quadratic

equation ax^{2}+bx+c = 0. This is done without actually calculating the
solutions.

Consider the quadratic equation

x^{2} − 4x + 4 = 0.

Calculate the discriminant and use it to determine the nature and number of the

solutions.

Compare x^{2} − 4x + 4 = 0 with ax^{2} + bx + c = 0 and note
that a = 1, b = −4, and

c = 4. The discriminant is given by the calculation

D = b^{2} − 4ac = (−4)^{2} − 4(1)(4) = 0.

Note that the discriminant equals zero.

Consider the quadratic function f(x) = x^{2} −4x+4, which can be written
in vertex

form

f(x) = (x − 2)^{2}. (17)

This is a parabola that opens upward and is shifted 2 units to the right. Note
that

there is no vertical shift, so the vertex of the parabola will rest on the
x-axis, as shown

in** Figure 3**. In this case, we found it necessary to plot two points to
the right of the

axis of symmetry, then mirror them across the axis of symmetry, in order to get
an

accurate plot of the parabola.

Figure 3. At the right is a table of points satisfying
f(x) = (x−2)^{2}. These

points and their mirror images are seen as solid dots superimposed on the

graph of f(x) = (x − 2)^{2} at the left.

Take a closer look at **equation (17).** If we set f(x) = 0 in this equation,
then we

get 0 = (x − 2)^{2}. This could be written 0 = (x − 2)(x − 2) and we
could say that the

solutions are 2 and 2 again. However, mathematicians prefer to say that “2 is a
solution

of multiplicity 2” or “2 is a double solution.”Note how the parabola is tangent
to

the x-axis at the location of the “double solution.” That is, the parabola comes
down

from positive infinity, touches (but does not cross) the x-axis at x = 2, then
rises again

to positive infinity. Of course, the situation would be reversed in the parabola
opened

downward, as in g(x) = −(x − 2)^{2}, but the graph would still “kiss”
the x-axis at the

location of the “double solution.”

Still, the key thing to note here is the fact that the discriminant D = 0 and
the

parabola has only one x-intercept. That is, the equation x^{2} − 4x + 4
= 0 has a single

real solution.

Next, let’s look what happens when the discriminant is negative.

Example 18. Consider the quadratic equation

x^{2} − 4x + 8 = 0.

Calculate the discriminant and use it to determine the nature and number of the

solutions.

Compare x^{2} − 4x + 8 = 0 with ax^{2} + bx + c = 0 and note
that a = 1, b = −4, and

c = 8. The discriminant is given by the calculation

D = b^{2} − 4ac = (−4)^{2} − 4(1)(8) = −16.

Note that the discriminant is negative.

Consider the quadratic function f(x) = x^{2} −4x+8, which can be written
in vertex

form

f(x) = (x − 2)^{2} + 4.

This is a parabola that opens upward. Moreover, it has to be shifted 2 units to
the right

and 4 units upward, so there can be no x-intercepts, as shown in **Figure 4.**
Again,

we found it necessary in this example to plot two points to the right of the
axis of

symmetry, then mirror them, in order to get an accurate plot of the parabola.

Figure 4. At the right is a table of points satisfying
f(x) = (x − 2)^{2} + 4.

These points and their mirror images are seen as solid dots superimposed on

the graph of f(x) = (x − 2)^{2} + 4 at the left.

Once again, the key point in this example is the fact that the discriminant is
negative

and there are no real solutions of the quadratic equation (equivalently, there
are no

x-intercepts). Let’s see what happens if we actually try to find the solutions
of x^{2} −

4x + 8 = 0 using the quadratic formula. Again, a = 1, b = −4, and c = 8, so

Simplifying ,

Again, remember that the number under the square root is the discriminant . In
this

case the disriminant is −16. It is not possible to square a real number and get
−16.

Thus, the quadratic equation x^{2} − 4x + 8 = 0 has no real solutions,
as predicted.

Let’s summarize the findings in our last three examples.

**Summary 19**. Consider the quadratic equation

ax

^{2}+ bx + c = 0.

The discriminant is
defined as

D = b^{2} − 4ac.

There are three possibilities:

1. If D > 0, then the quadratic equation has two real solutions.

2. If D = 0, then the quadratic equation has one real solution.

3. If D < 0, then the quadratic equation has no real solutions.

This key result is reflected in the graph of the quadratic function.

**Summary 20**. Consider the quadratic function

f(x) = ax

^{2}+ bx + c.

The graph of
this function is a parabola. Three possibilities exist depending upon

the value of the discriminant D = b^{2} − 4ac.

1. If D > 0, the parabola has two x-intercepts.

2. If D = 0, the parabola has exactly one x-intercept.

3. If D < 0, the parabola has no x-intercepts.

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