Question

Solve, graph and give interval notation
x²-5x/x-3 ≥0

Answer 1

To solve the inequality \(\frac{x^2 - 5x}{x - 3} \geq 0\), we first need to identify the points where the expression is equal to zero or undefined.

**Step 1: Find the zeros of the numerator and the points where the expression is undefined.**

1. **Numerator:** \(x^2 - 5x = 0\)

Factor out \(x\):
\[
x(x - 5) = 0
\]
This gives us the solutions:
\[
x = 0 \quad \text{and} \quad x = 5
\]

2. **Denominator:** \(x - 3 = 0\)

This gives us:
\[
x = 3
\]

So, the critical points are \(x = 0\), \(x = 3\), and \(x = 5\).

**Step 2: Determine the sign of the expression in each interval defined by these critical points.**

The critical points divide the number line into the following intervals:
- \((-\infty, 0)\)
- \((0, 3)\)
- \((3, 5)\)
- \((5, \infty)\)

Now, we will test points from each interval:

1. **Interval \((-\infty, 0)\):** Choose \(x = -1\):
\[
\frac{(-1)^2 - 5(-1)}{-1 - 3} = \frac{1 + 5}{-4} = \frac{6}{-4} < 0
\]

2. **Interval \((0, 3)\):** Choose \(x = 1\):
\[
\frac{1^2 - 5(1)}{1 - 3} = \frac{1 - 5}{-2} = \frac{-4}{-2} = 2 > 0
\]

3. **Interval \((3, 5)\):** Choose \(x = 4\):
\[
\frac{4^2 - 5(4)}{4 - 3} = \frac{16 - 20}{1} = -4 < 0
\]

4. **Interval \((5, \infty)\):** Choose \(x = 6\):
\[
\frac{6^2 - 5(6)}{6 - 3} = \frac{36 - 30}{3} = \frac{6}{3} = 2 > 0
\]

**Step 3: Summarize the results:**

- The expression is \(\leq 0\) in the intervals \((-\infty, 0)\) and \((3, 5)\).
- It is \(0\) at \(x = 0\) and \(x = 5\).
- The expression is undefined at \(x = 3\).

**Step 4: Write the solution in interval notation:**

Since we want \(\frac{x^2 - 5x}{x - 3} \geq 0\), we include \(x = 0\) and \(x = 5\) but exclude \(x = 3\):
\[
\text{Solution: } [0, 3) \cup [5, \infty)
\]

**Step 5: Graph the solution:**

On a number line, you can represent the solution as follows:
- Draw a solid dot at \(0\) and \(5\) indicating inclusive endpoints.
- Draw an open circle at \(3\) indicating it is not included in the solution.
- Shade between \(0\) and \(3\), and from \(5\) to \(\infty\).

Thus, the solution to the inequality \(\frac{x^2 - 5x}{x - 3} \geq 0\) is:

\[
\boxed{[0, 3) \cup [5, \infty)}
\]