If `(13x^(2) + 30 x + 17)/(3x^(2) + 14 x + 8) gt 3`, then find the value of x, if `x in R` and `x ne (-2)/(3), -4`.

Options

A. `(-oo, -4) cup ((-2)/(3), (7)/(2)) cup ((9)/(2), oo)`
B. `(-oo, -4) cup ((-2)/(3), (-1)/(2)) cup ((7)/(2), oo)`
C. `(-4, (-2)/(3)) cup ((-1)/(2), (7)/(2))`
D. None of these

Answer

To solve the inequality \(\frac{13x^2 + 30x + 17}{3x^2 + 14x + 8} > 3\), we can follow these steps:

### Step 1: Rewrite the Inequality
We start with the inequality:
\[ \frac{13x^2 + 30x + 17}{3x^2 + 14x + 8} > 3 \]
Subtract 3 from both sides:
\[ \frac{13x^2 + 30x + 17}{3x^2 + 14x + 8} - 3 > 0 \]

### Step 2: Combine the Terms
To combine the terms, we need a common denominator:
\[ \frac{13x^2 + 30x + 17 - 3(3x^2 + 14x + 8)}{3x^2 + 14x + 8} > 0 \]
This simplifies to:
\[ \frac{13x^2 + 30x + 17 - (9x^2 + 42x + 24)}{3x^2 + 14x + 8} > 0 \]
Combining the terms in the numerator:
\[ \frac{(13x^2 - 9x^2) + (30x - 42x) + (17 - 24)}{3x^2 + 14x + 8} > 0 \]
This gives:
\[ \frac{4x^2 - 12x - 7}{3x^2 + 14x + 8} > 0 \]

### Step 3: Factor the Numerator and Denominator
Next, we need to factor the numerator \(4x^2 - 12x - 7\) and the denominator \(3x^2 + 14x + 8\).

1. **Numerator**:
We can use the quadratic formula to find the roots:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} = \frac{12 \pm \sqrt{(-12)^2 - 4 \cdot 4 \cdot (-7)}}{2 \cdot 4} \]
\[ = \frac{12 \pm \sqrt{144 + 112}}{8} = \frac{12 \pm \sqrt{256}}{8} = \frac{12 \pm 16}{8} \]
This gives us the roots:
\[ x = \frac{28}{8} = 3.5 \quad \text{and} \quad x = \frac{-4}{8} = -0.5 \]
Thus, the numerator factors to:
\[ 4(x - 3.5)(x + 0.5) \]

2. **Denominator**:
We can also use the quadratic formula for the denominator:
\[ x = \frac{-14 \pm \sqrt{14^2 - 4 \cdot 3 \cdot 8}}{2 \cdot 3} = \frac{-14 \pm \sqrt{196 - 96}}{6} = \frac{-14 \pm \sqrt{100}}{6} \]
This gives us:
\[ x = \frac{-14 \pm 10}{6} \]
Thus, the roots are:
\[ x = \frac{-4}{6} = -\frac{2}{3} \quad \text{and} \quad x = \frac{-24}{6} = -4 \]
The denominator factors to:
\[ 3(x + 4)(x + \frac{2}{3}) \]

### Step 4: Set Up the Inequality
Now we can rewrite the inequality:
\[ \frac{4(x - 3.5)(x + 0.5)}{3(x + 4)(x + \frac{2}{3})} > 0 \]

### Step 5: Find Critical Points
The critical points are:
- From the numerator: \(x = 3.5\) and \(x = -0.5\)
- From the denominator: \(x = -4\) and \(x = -\frac{2}{3}\)

### Step 6: Test Intervals
We will test the sign of the expression in the intervals defined by the critical points:
1. \( (-\infty, -4) \)
2. \( (-4, -\frac{2}{3}) \)
3. \( (-\frac{2}{3}, -0.5) \)
4. \( (-0.5, 3.5) \)
5. \( (3.5, \infty) \)

### Step 7: Determine Where the Expression is Positive
After testing these intervals, we find that the expression is positive in the following intervals:
- \( (-\infty, -4) \)
- \( (-\frac{2}{3}, -0.5) \)
- \( (3.5, \infty) \)

### Step 8: Write the Final Solution
Thus, the solution to the inequality is:
\[ x \in (-\infty, -4) \cup \left(-\frac{2}{3}, -0.5\right) \cup (3.5, \infty) \]
with the restrictions that \(x \neq -\frac{2}{3}\) and \(x \neq -4\).