Question

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

(ii) x²+3x>10

(iii) 9-2|4x+1|>3

Answer 1

Let's solve each inequality step by step:

### (i) \( \frac{x^2 - 5x}{x - 3} \geq 0 \)

**Step 1: Find critical points.**

The expression can be simplified as follows:

- The numerator is \( x^2 - 5x = x(x - 5) \).
- The critical points occur when the numerator is zero or the denominator is undefined.

Setting the numerator to zero:
1. \( x(x - 5) = 0 \) gives \( x = 0 \) and \( x = 5 \).

Setting the denominator to zero:
2. \( x - 3 = 0 \) gives \( x = 3 \).

**Step 2: Analyze the intervals.**

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

**Step 3: Test a point in each interval to determine the sign.**

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 4: Combine results and include critical points in the solution.**

- \( (-\infty, 0) \): negative
- \( (0, 3) \): positive
- \( (3, 5) \): negative
- \( (5, \infty) \): positive

At critical points \( 0 \) and \( 5 \), the expression is zero and included in the solution:
The value is undefined at \( x = 3 \).

**Solution:**

In interval notation:
\[
[0, 3) \cup [5, \infty)
\]

---

### (ii) \( x^2 + 3x > 10 \)

**Step 1: Rewrite the inequality.**
\[
x^2 + 3x - 10 > 0
\]

**Step 2: Factor the left-hand side.**
\[
(x + 5)(x - 2) > 0
\]

**Step 3: Identify the critical points.**
The roots are \( x = -5 \) and \( x = 2 \).

**Step 4: Test intervals:**
The intervals are \( (-\infty, -5), (-5, 2), (2, \infty) \).

1. **Interval \( (-\infty, -5) \)**: Choose \( x = -6 \)
\[
(-6 + 5)(-6 - 2) = (-1)(-8) = 8 > 0
\]

2. **Interval \( (-5, 2) \)**: Choose \( x = 0 \)
\[
(0 + 5)(0 - 2) = (5)(-2) = -10 < 0
\]

3. **Interval \( (2, \infty) \)**: Choose \( x = 3 \)
\[
(3 + 5)(3 - 2) = (8)(1) = 8 > 0
\]

**Solution:**

In interval notation:
\[
(-\infty, -5) \cup (2, \infty)
\]

---

### (iii) \( 9 - 2|4x + 1| > 3 \)

**Step 1: Isolate the absolute value.**
\[
9 - 3 > 2 |4x + 1| \rightarrow 6 > 2 |4x + 1| \rightarrow 3 > |4x + 1|
\]

**Step 2: Rewrite the absolute value inequality.**
\[
-3 < 4x + 1 < 3
\]

**Step 3: Solve the compound inequality.**

1. For \( 4x + 1 < 3 \):
\[
4x < 2 \rightarrow x < \frac{2}{4} \rightarrow x < \frac{1}{2}
\]

2. For \( -3 < 4x + 1 \):
\[
-4 < 4x \rightarrow -1 < x
\]

**Combine results**:
\[
-1 < x < \frac{1}{2}
\]

**Solution:**

In interval notation:
\[
(-1, \frac{1}{2})
\]

In summary, the solutions are:

(i) \( [0, 3) \cup [5, \infty) \)

(ii) \( (-\infty, -5) \cup (2, \infty) \)

(iii) \( (-1, \frac{1}{2}) \)