5.4 The Quadratic Formula - College of the Redwoods

494 Chapter 5 Quadratic Functions
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
Note that the discriminant is negative.
D = b 2 − 4ac = (−4) 2 − 4(1)(8) = −16.
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.
y
10
(2, 4) x f(x) = (x−2) 2 +4
x
10
3 5
4 8
x = 2
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
x = −b ± √ b 2 − 4ac
2a
= −(−4) ± √ (−4) 2 − 4(1)(8)
.
2(1)
Simplifying,
x = 4 ± √ −16
.
2
Version: Fall 2007