Question

Solve, graph and give interval notation for the inequality
(i) x²-5x/x-3≥0

Answer 1

To solve the inequality \(\frac{x^2 - 5x}{x - 3} \geq 0\), we will follow these steps:

1. **Factor the numerator:**
The numerator \(x^2 - 5x\) can be factored as:
\[
x(x - 5)
\]
Therefore, the inequality becomes:
\[
\frac{x(x - 5)}{x - 3} \geq 0
\]

2. **Identify critical points:**
The critical points occur where the numerator is zero or where the denominator is zero.

- Set the numerator equal to zero:
\[
x(x - 5) = 0 \implies x = 0 \text{ or } x = 5
\]
- Set the denominator equal to zero:
\[
x - 3 = 0 \implies x = 3
\]

Therefore, the critical points are:
- \(x = 0\)
- \(x = 3\)
- \(x = 5\)

3. **Determine test intervals:**
The critical points divide the number line into four intervals:
- \((-\infty, 0)\)
- \((0, 3)\)
- \((3, 5)\)
- \((5, \infty)\)

4. **Test each interval:**
Choose a test point in each interval to determine the sign of the expression \(\frac{x(x - 5)}{x - 3}\).

- For the interval \((-\infty, 0)\), use \(x = -1\):
\[
\frac{(-1)(-1 - 5)}{-1 - 3} = \frac{-1 \cdot -6}{-4} = \frac{6}{4} > 0
\]
- For the interval \((0, 3)\), use \(x = 1\):
\[
\frac{(1)(1 - 5)}{1 - 3} = \frac{1 \cdot -4}{-2} = \frac{-4}{-2} = 2 > 0
\]
- For the interval \((3, 5)\), use \(x = 4\):
\[
\frac{(4)(4 - 5)}{4 - 3} = \frac{4 \cdot -1}{1} = -4 < 0
\]
- For the interval \((5, \infty)\), use \(x = 6\):
\[
\frac{(6)(6 - 5)}{6 - 3} = \frac{6 \cdot 1}{3} = 2 > 0
\]

5. **Combine results:**
The inequality \(\frac{x(x - 5)}{x - 3} \geq 0\) is satisfied in the intervals where the expression is positive or zero.

- Positive in \((-\infty, 0)\), \((0, 3)\), and \((5, \infty)\).
- Zero at \(x = 0\) and \(x = 5\).

The expression is undefined at \(x = 3\).

6. **Write the solution in interval notation:**
Therefore, the solution to the inequality in interval notation is:
\[
(-\infty, 0] \cup (0, 3) \cup [5, \infty)
\]

7. **Graph the solution:**
To graph, plot the points \(0\), \(3\), and \(5\):

- Use a closed circle at \(0\) and \(5\) to indicate that those points are included in the solution.
- Use an open circle at \(3\) to indicate that it is not included.
- Shade the regions that satisfy the inequality.

Putting this together, the final answer is:

**Solution in Interval Notation:**
\[
(-\infty, 0] \cup (0, 3) \cup [5, \infty)
\]