Consider the inequality
\(x^2 - 6x + 8 \gt 0\text{.}\) Similar to before, first find the critical values, by factoring,
\begin{align*}
x^2 - 6x + 8 \amp= 0\\
(x-2)(x-4) \amp= 0
\end{align*}
So, the critical values are
\(x = 2\) and
\(x = 4\text{.}\) Set up a number line, with these roots marked,
These roots split up the number line into 3 intervals:
\(x \lt 2\text{,}\) \(2 \lt x \lt 4\text{,}\) and
\(x \gt 4\text{.}\) For each of these intervals, the quadratic will either be positive everywhere, or negative everywhere. So, it is enough to test a single point, and see if the quadratic is positive or negative. For example,
-
For
\(x \lt 2\text{,}\) choose
\(x = 1\text{:}\)
\begin{equation*}
(1 - 2)(1 - 4) = (-1)(-3) = 3 \gt 0 \quad \text{(positive)}
\end{equation*}
-
For
\(2 \lt x \lt 4\text{,}\) choose
\(x = 3\text{:}\)
\begin{equation*}
(3 - 2)(3 - 4) = (1)(-1) = -1 \lt 0 \quad \text{(negative)}
\end{equation*}
-
For
\(x \gt 4\text{,}\) choose
\(x = 5\text{:}\)
\begin{equation*}
(5 - 2)(5 - 4) = (3)(1) = 3 \gt 0 \quad \text{(positive)}
\end{equation*}
To illustrate these results, we can put a
\(+\) or
\(-\) above each interval, depending if it’s positive or negative.
This means that the solution is
\(x \lt 2\) or
\(x \gt 4\text{.}\) We can denote this graphically by highlighting those regions.
We use open circles (basically, hollow circles) to indicate that
\(x=2\) and
\(x=4\) are not included in the solution, because the inequality is
\(\lt\text{.}\)
In fact, when doing test points, we only need to know the sign of the output (whether it is positive or negative). Its exact value as a number isn’t necessary. Recall that with multiplying positive and negative numbers,
\begin{align*}
(+)(+) \amp= +\\
(+)(-) \amp= - \quad \text{and} \quad (-)(+) = -\\
(-)(-) \amp= +
\end{align*}
Then, testing points could look like this,
-
Test
\(x = 1\text{:}\) \((1 - 2)(1 - 4) = (-)(-) = +\)
-
Test
\(x = 3\text{:}\) \((3 - 2)(3 - 4) = (+)(-) = -\)
-
Test
\(x = 5\text{:}\) \((5 - 2)(5 - 4) = (+)(+) = +\)
