If `(x^2+2x+7)/(2x+3)<6`, x`in` R then

Options

A. `x in (-oo, -(3)/(2)) uu (11, oo)`
B. `x in (-oo, -1) uu ( 11, oo)`
C. `x in (-(3)/(2), -1)`
D. `x in (-oo, -(3)/(2)) uu (-1, 11)`

Answer

To solve the inequality \(\frac{x^2 + 2x + 7}{2x + 3} < 6\), we will follow these steps:

### Step 1: Rewrite the Inequality
We start by rewriting the inequality:
\[ \frac{x^2 + 2x + 7}{2x + 3} - 6 < 0 \]
This simplifies to:
\[ \frac{x^2 + 2x + 7 - 6(2x + 3)}{2x + 3} < 0 \]

### Step 2: Simplify the Numerator
Now, we simplify the numerator:
\[ x^2 + 2x + 7 - 12x - 18 = x^2 - 10x - 11 \]
Thus, the inequality becomes:
\[ \frac{x^2 - 10x - 11}{2x + 3} < 0 \]

### Step 3: Factor the Numerator
Next, we factor the quadratic expression in the numerator:
\[ x^2 - 10x - 11 = (x - 11)(x + 1) \]
So, we rewrite the inequality as:
\[ \frac{(x - 11)(x + 1)}{2x + 3} < 0 \]

### Step 4: Find Critical Points
We need to find the critical points where the expression is equal to zero or undefined:
1. \(x - 11 = 0 \Rightarrow x = 11\)
2. \(x + 1 = 0 \Rightarrow x = -1\)
3. \(2x + 3 = 0 \Rightarrow x = -\frac{3}{2}\)

The critical points are \(x = -1\), \(x = 11\), and \(x = -\frac{3}{2}\).

### Step 5: Test Intervals
We will test the sign of the expression in the intervals defined by these critical points:
- Interval 1: \((-∞, -\frac{3}{2})\)
- Interval 2: \((- \frac{3}{2}, -1)\)
- Interval 3: \((-1, 11)\)
- Interval 4: \((11, ∞)\)

**Testing Interval 1: Choose \(x = -2\)**
\[ \frac{(-2 - 11)(-2 + 1)}{2(-2) + 3} = \frac{(-13)(-1)}{-1} = 13 > 0 \]

**Testing Interval 2: Choose \(x = -1.5\)**
\[ \frac{(-1.5 - 11)(-1.5 + 1)}{2(-1.5) + 3} = \frac{(-12.5)(-0.5)}{0} \text{ (undefined)} \]

**Testing Interval 3: Choose \(x = 0\)**
\[ \frac{(0 - 11)(0 + 1)}{2(0) + 3} = \frac{(-11)(1)}{3} = -\frac{11}{3} < 0 \]

**Testing Interval 4: Choose \(x = 12\)**
\[ \frac{(12 - 11)(12 + 1)}{2(12) + 3} = \frac{(1)(13)}{27} > 0 \]

### Step 6: Determine the Solution
From our tests, we find:
- The expression is positive in \((-∞, -\frac{3}{2})\) and \((11, ∞)\).
- The expression is negative in \((-1, 11)\).

Thus, the solution to the inequality \(\frac{(x - 11)(x + 1)}{2x + 3} < 0\) is:
\[ x \in (-\frac{3}{2}, -1) \cup (-1, 11) \]

### Final Answer
The solution set is:
\[ x \in (-\infty, -\frac{3}{2}) \cup (-1, 11) \]

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