Number of intergal values of x satisfying the inequality `(x^2+6x-7)/(|x+2||x+3|) lt 0 `is

Options

A. 5
B. 6
C. 7
D. 8

Explanation

To solve the inequality \(\frac{x^2 + 6x - 7}{|x + 2||x + 3|} < 0\), we will follow these steps:

### Step 1: Identify the points where the expression is undefined
The expression is undefined when the denominator is zero. Thus, we need to find when \(|x + 2| = 0\) and \(|x + 3| = 0\).

- \(|x + 2| = 0 \Rightarrow x + 2 = 0 \Rightarrow x = -2\)
- \(|x + 3| = 0 \Rightarrow x + 3 = 0 \Rightarrow x = -3\)

The expression is undefined at \(x = -2\) and \(x = -3\).

### Step 2: Analyze the numerator
Next, we analyze the numerator \(x^2 + 6x - 7\). We will find the roots of the quadratic equation by factoring or using the quadratic formula.

The quadratic can be factored as follows:
\[ x^2 + 6x - 7 = (x + 7)(x - 1) \]
Thus, the roots are \(x = -7\) and \(x = 1\).

### Step 3: Determine the intervals
The critical points from the numerator and the points where the expression is undefined give us the intervals to test:
- \(x = -7\)
- \(x = -3\)
- \(x = -2\)
- \(x = 1\)

This divides the real line into the following intervals:
1. \((- \infty, -7)\)
2. \((-7, -3)\)
3. \((-3, -2)\)
4. \((-2, 1)\)
5. \((1, \infty)\)

### Step 4: Test each interval
We will test a point from each interval to determine if the inequality holds.

1. **Interval \((- \infty, -7)\)**: Choose \(x = -8\)
\[ \frac{(-8)^2 + 6(-8) - 7}{|(-8) + 2||(-8) + 3|} = \frac{64 - 48 - 7}{6 \cdot 5} = \frac{9}{30} > 0 \]

2. **Interval \((-7, -3)\)**: Choose \(x = -5\)
\[ \frac{(-5)^2 + 6(-5) - 7}{|(-5) + 2||(-5) + 3|} = \frac{25 - 30 - 7}{3 \cdot 2} = \frac{-12}{6} < 0 \]

3. **Interval \((-3, -2)\)**: Choose \(x = -2.5\)
\[ \frac{(-2.5)^2 + 6(-2.5) - 7}{|(-2.5) + 2||(-2.5) + 3|} = \frac{6.25 - 15 - 7}{0.5 \cdot 0.5} = \frac{-15.75}{0.25} < 0 \]

4. **Interval \((-2, 1)\)**: Choose \(x = 0\)
\[ \frac{0^2 + 6(0) - 7}{|0 + 2||0 + 3|} = \frac{-7}{2 \cdot 3} = \frac{-7}{6} < 0 \]

5. **Interval \((1, \infty)\)**: Choose \(x = 2\)
\[ \frac{(2)^2 + 6(2) - 7}{|2 + 2||2 + 3|} = \frac{4 + 12 - 7}{4 \cdot 5} = \frac{9}{20} > 0 \]

### Step 5: Combine results
From our tests, the inequality \(\frac{x^2 + 6x - 7}{|x + 2||x + 3|} < 0\) holds in the intervals:
- \((-7, -3)\)
- \((-3, -2)\)
- \((-2, 1)\)

### Step 6: Identify integral values
Now we find the integral values in these intervals:
- In \((-7, -3)\): \(-6, -5, -4\) (3 values)
- In \((-3, -2)\): No integral values
- In \((-2, 1)\): \(-1, 0\) (2 values)

### Conclusion
The total number of integral values satisfying the inequality is:
\[ 3 + 0 + 2 = 5 \]

Thus, the answer is **5**.

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