`(8x^(2) + 16x - 51)/((2x-3)(x+4)) gt 3`, if x satisfies

Options

A. `x lt - 4`
B. `-3 lt x lt 3//2`
C. `x gt 5//2`
D. All of these

Answer

To solve the inequality \((8x^{2} + 16x - 51)/((2x-3)(x+4)) > 3\), we will follow these steps:

### Step 1: Rearrange the Inequality
We start by moving all terms to one side of the inequality:
\[ \frac{8x^{2} + 16x - 51}{(2x-3)(x+4)} - 3 > 0 \]

### Step 2: Combine the Terms
To combine the terms, we need a common denominator:
\[ \frac{8x^{2} + 16x - 51 - 3(2x-3)(x+4)}{(2x-3)(x+4)} > 0 \]

### Step 3: Expand the Denominator
Now, we expand \(3(2x-3)(x+4)\):
\[ 3(2x^2 + 8x - 3x - 12) = 3(2x^2 + 5x - 12) = 6x^2 + 15x - 36 \]

### Step 4: Substitute Back
Now substitute back into the inequality:
\[ \frac{8x^{2} + 16x - 51 - (6x^{2} + 15x - 36)}{(2x-3)(x+4)} > 0 \]

### Step 5: Simplify the Numerator
Now simplify the numerator:
\[ 8x^{2} + 16x - 51 - 6x^{2} - 15x + 36 = 2x^{2} + x - 15 \]

### Step 6: Factor the Quadratic
Next, we need to factor \(2x^{2} + x - 15\). We look for two numbers that multiply to \(2 \times -15 = -30\) and add to \(1\). These numbers are \(6\) and \(-5\). Thus, we can factor:
\[ 2x^{2} + 6x - 5x - 15 = (2x - 5)(x + 3) \]

### Step 7: Set Up the Inequality
Now we can rewrite the inequality:
\[ \frac{(2x - 5)(x + 3)}{(2x - 3)(x + 4)} > 0 \]

### Step 8: Find Critical Points
The critical points are found by setting the numerator and denominator to zero:
1. \(2x - 5 = 0 \Rightarrow x = \frac{5}{2}\)
2. \(x + 3 = 0 \Rightarrow x = -3\)
3. \(2x - 3 = 0 \Rightarrow x = \frac{3}{2}\)
4. \(x + 4 = 0 \Rightarrow x = -4\)

### Step 9: Test Intervals
We will test the intervals defined by the critical points: \((-∞, -4)\), \((-4, -3)\), \((-3, \frac{3}{2})\), \((\frac{3}{2}, \frac{5}{2})\), and \((\frac{5}{2}, ∞)\).

1. **Interval \((-∞, -4)\)**: Choose \(x = -5\) → positive.
2. **Interval \((-4, -3)\)**: Choose \(x = -3.5\) → negative.
3. **Interval \((-3, \frac{3}{2})\)**: Choose \(x = 0\) → negative.
4. **Interval \((\frac{3}{2}, \frac{5}{2})\)**: Choose \(x = 2\) → positive.
5. **Interval \((\frac{5}{2}, ∞)\)**: Choose \(x = 3\) → positive.

### Step 10: Write the Solution
The solution to the inequality is where the expression is positive:
\[ x \in (-∞, -4) \cup \left(\frac{3}{2}, \frac{5}{2}\right) \]

### Final Step: Check the Options
Now we check the options provided in the question against our solution. The only valid option is \(x < -4\).