To solve and graph the solution to the inequality [tex]\( x^2 - 10x > -24 \)[/tex] on the number line, follow these steps:
1. Rewrite the inequality:
[tex]\[
x^2 - 10x + 24 > 0
\][/tex]
This is achieved by adding 24 to both sides of the original inequality.
2. Find the roots of the corresponding equality:
To find the points where the expression equals zero, solve the quadratic equation:
[tex]\[
x^2 - 10x + 24 = 0
\][/tex]
Factorize the equation:
[tex]\[
(x - 4)(x - 6) = 0
\][/tex]
Therefore, the roots are:
[tex]\[
x = 4 \quad \text{and} \quad x = 6
\][/tex]
3. Analyze the intervals:
Because the quadratic [tex]\( x^2 - 10x + 24 \)[/tex] changes sign at the roots [tex]\( x = 4 \)[/tex] and [tex]\( x = 6 \)[/tex], divide the real line into intervals using these points:
[tex]\[
(-\infty, 4), \quad (4, 6), \quad (6, \infty)
\][/tex]
4. Determine the sign of the expression in each interval:
- For [tex]\( x \in (-\infty, 4) \)[/tex]: Choose a test point like [tex]\( x = 0 \)[/tex]. Substituting [tex]\( x = 0 \)[/tex] into the expression:
[tex]\[
0^2 - 10(0) + 24 = 24 > 0
\][/tex]
Thus, the inequality [tex]\( x^2 - 10x + 24 > 0 \)[/tex] is satisfied in [tex]\( (-\infty, 4) \)[/tex].
- For [tex]\( x \in (4, 6) \)[/tex]: Choose a test point like [tex]\( x = 5 \)[/tex]. Substituting [tex]\( x = 5 \)[/tex] into the expression:
[tex]\[
5^2 - 10(5) + 24 = -1 < 0
\][/tex]
Thus, the inequality [tex]\( x^2 - 10x + 24 > 0 \)[/tex] is not satisfied in [tex]\( (4, 6) \)[/tex].
- For [tex]\( x \in (6, \infty) \)[/tex]: Choose a test point like [tex]\( x = 7 \)[/tex]. Substituting [tex]\( x = 7 \)[/tex] into the expression:
[tex]\[
7^2 - 10(7) + 24 = 49 - 70 + 24 = 3 > 0
\][/tex]
Thus, the inequality [tex]\( x^2 - 10x + 24 > 0 \)[/tex] is satisfied in [tex]\( (6, \infty) \)[/tex].
5. Combine the intervals:
The solution to the inequality [tex]\( x^2 - 10x + 24 > 0 \)[/tex] is:
[tex]\[
x \in (-\infty, 4) \cup (6, \infty)
\][/tex]
6. Graph the solution on the number line:
- Draw a number line.
- Use an open circle at [tex]\( x = 4 \)[/tex] to indicate that 4 is not included in the solution.
- Shade the region to the left of 4, towards [tex]\( -\infty \)[/tex].
- Use an open circle at [tex]\( x = 6 \)[/tex] to indicate that 6 is not included in the solution.
- Shade the region to the right of 6, towards [tex]\( \infty \)[/tex].
The final graph of the solution on the number line looks like this:
[tex]\[
\begin{array}{c}
\text{Number Line} \\
\begin{tikzpicture}
\draw[thick] (-6.5, 0) -- (8.5, 0);
\foreach \x in {-6, -2, 2, 6} \draw (\x, 0.1) -- (\x, -0.1);
\node at (-0.2, -0.4) {0};
\node[circle,fill=white,draw,inner sep=2pt] at (4,0) {};
\node[below] at (4, -0.4) {4};
\node[circle,fill=white,draw,inner sep=2pt] at (6,0) {};
\node[below] at (6, -0.4) {6};
\draw[->, thick] (-6.5, 0) -- (4,0);
\draw[->, thick] (6,0) -- (8.5,0);
\end{tikzpicture}
\end{array}
\][/tex]
