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To solve the inequality [tex]\(\frac{x^2 + x - 6}{x - 7} \leq 0\)[/tex], let's go through it step-by-step.
### Step 1: Factorize the numerator
First, we factorize the quadratic expression in the numerator:
[tex]\[ x^2 + x - 6 \][/tex]
We look for two numbers that multiply to [tex]\(-6\)[/tex] and add to [tex]\(1\)[/tex]. These numbers are [tex]\(3\)[/tex] and [tex]\(-2\)[/tex]. Thus,
[tex]\[ x^2 + x - 6 = (x + 3)(x - 2) \][/tex]
### Step 2: Rewrite the inequality
Substitute the factorized form of the quadratic expression back into the inequality:
[tex]\[ \frac{(x + 3)(x - 2)}{x - 7} \leq 0 \][/tex]
### Step 3: Determine critical points
The critical points are where the numerator and denominator are zero:
1. [tex]\(x + 3 = 0 \implies x = -3\)[/tex]
2. [tex]\(x - 2 = 0 \implies x = 2\)[/tex]
3. [tex]\(x - 7 = 0 \implies x = 7\)[/tex]
### Step 4: Sign analysis
We need to analyze the signs of each factor in the intervals determined by the critical points:
- Intervals: [tex]\((- \infty, -3]\)[/tex], [tex]\((-3, 2]\)[/tex], [tex]\((2, 7)\)[/tex], [tex]\( (7, \infty) \)[/tex]
Check the sign of [tex]\(\frac{(x + 3)(x - 2)}{x - 7}\)[/tex] in each interval:
1. For [tex]\(x \in (- \infty, -3)\)[/tex]:
- [tex]\(x + 3 < 0\)[/tex]
- [tex]\(x - 2 < 0\)[/tex]
- [tex]\(x - 7 < 0\)[/tex]
The expression is [tex]\(\frac{(-)(-)}{-} = \frac{+}{-} = -\)[/tex]. Negative.
2. For [tex]\(x \in [-3, 2]\)[/tex]:
- [tex]\(x + 3 \geq 0\)[/tex]
- [tex]\(x - 2 \leq 0\)[/tex]
- [tex]\(x - 7 < 0\)[/tex]
For [tex]\(x = -3\)[/tex]: [tex]\(\frac{0 \cdot (-5)}{-10} = 0\)[/tex], satisfies because [tex]\(\leq 0\)[/tex].
For [tex]\(x \in (-3, 2)\)[/tex]: The expression [tex]\(\frac{(+)(-)}{-} = \frac{+}{-} = -\)[/tex]. Negative.
For [tex]\(x = 2\)[/tex]: [tex]\(\frac{- \cdot 0}{-5} = 0\)[/tex], satisfies because [tex]\(\leq 0\)[/tex].
3. For [tex]\(x \in (2, 7)\)[/tex]:
- [tex]\(x + 3 > 0\)[/tex]
- [tex]\(x - 2 > 0\)[/tex]
- [tex]\(x - 7 < 0\)[/tex]
The expression is [tex]\(\frac{(+)(+)}{-} = \frac{+}{-} = -\)[/tex]. Negative.
4. For [tex]\(x \in (7, \infty)\)[/tex]:
- [tex]\(x + 3 > 0\)[/tex]
- [tex]\(x - 2 > 0\)[/tex]
- [tex]\(x - 7 > 0\)[/tex]
The expression [tex]\(\frac{(+)(+)}{+} = +\)[/tex]. Positive, not satisfying.
### Step 5: Combining intervals
The solution includes the points and intervals where the expression is less than or equal to zero:
[tex]\[ (- \infty, -3] \quad \text{and} \quad [2, 7) \][/tex]
However, we must include the endpoints [tex]\(-3\)[/tex] and [tex]\(2\)[/tex] but exclude [tex]\(7\)[/tex] since division by zero is not allowed.
### Final solution:
[tex]\[ x \leq -3 \quad \text{or} \quad 2 \leq x < 7 \][/tex]
Therefore, the correct interval is:
[tex]\[ \boxed{x \leq -3 \quad \text{or} \quad 2 \leq x < 7} \][/tex]
### Step 1: Factorize the numerator
First, we factorize the quadratic expression in the numerator:
[tex]\[ x^2 + x - 6 \][/tex]
We look for two numbers that multiply to [tex]\(-6\)[/tex] and add to [tex]\(1\)[/tex]. These numbers are [tex]\(3\)[/tex] and [tex]\(-2\)[/tex]. Thus,
[tex]\[ x^2 + x - 6 = (x + 3)(x - 2) \][/tex]
### Step 2: Rewrite the inequality
Substitute the factorized form of the quadratic expression back into the inequality:
[tex]\[ \frac{(x + 3)(x - 2)}{x - 7} \leq 0 \][/tex]
### Step 3: Determine critical points
The critical points are where the numerator and denominator are zero:
1. [tex]\(x + 3 = 0 \implies x = -3\)[/tex]
2. [tex]\(x - 2 = 0 \implies x = 2\)[/tex]
3. [tex]\(x - 7 = 0 \implies x = 7\)[/tex]
### Step 4: Sign analysis
We need to analyze the signs of each factor in the intervals determined by the critical points:
- Intervals: [tex]\((- \infty, -3]\)[/tex], [tex]\((-3, 2]\)[/tex], [tex]\((2, 7)\)[/tex], [tex]\( (7, \infty) \)[/tex]
Check the sign of [tex]\(\frac{(x + 3)(x - 2)}{x - 7}\)[/tex] in each interval:
1. For [tex]\(x \in (- \infty, -3)\)[/tex]:
- [tex]\(x + 3 < 0\)[/tex]
- [tex]\(x - 2 < 0\)[/tex]
- [tex]\(x - 7 < 0\)[/tex]
The expression is [tex]\(\frac{(-)(-)}{-} = \frac{+}{-} = -\)[/tex]. Negative.
2. For [tex]\(x \in [-3, 2]\)[/tex]:
- [tex]\(x + 3 \geq 0\)[/tex]
- [tex]\(x - 2 \leq 0\)[/tex]
- [tex]\(x - 7 < 0\)[/tex]
For [tex]\(x = -3\)[/tex]: [tex]\(\frac{0 \cdot (-5)}{-10} = 0\)[/tex], satisfies because [tex]\(\leq 0\)[/tex].
For [tex]\(x \in (-3, 2)\)[/tex]: The expression [tex]\(\frac{(+)(-)}{-} = \frac{+}{-} = -\)[/tex]. Negative.
For [tex]\(x = 2\)[/tex]: [tex]\(\frac{- \cdot 0}{-5} = 0\)[/tex], satisfies because [tex]\(\leq 0\)[/tex].
3. For [tex]\(x \in (2, 7)\)[/tex]:
- [tex]\(x + 3 > 0\)[/tex]
- [tex]\(x - 2 > 0\)[/tex]
- [tex]\(x - 7 < 0\)[/tex]
The expression is [tex]\(\frac{(+)(+)}{-} = \frac{+}{-} = -\)[/tex]. Negative.
4. For [tex]\(x \in (7, \infty)\)[/tex]:
- [tex]\(x + 3 > 0\)[/tex]
- [tex]\(x - 2 > 0\)[/tex]
- [tex]\(x - 7 > 0\)[/tex]
The expression [tex]\(\frac{(+)(+)}{+} = +\)[/tex]. Positive, not satisfying.
### Step 5: Combining intervals
The solution includes the points and intervals where the expression is less than or equal to zero:
[tex]\[ (- \infty, -3] \quad \text{and} \quad [2, 7) \][/tex]
However, we must include the endpoints [tex]\(-3\)[/tex] and [tex]\(2\)[/tex] but exclude [tex]\(7\)[/tex] since division by zero is not allowed.
### Final solution:
[tex]\[ x \leq -3 \quad \text{or} \quad 2 \leq x < 7 \][/tex]
Therefore, the correct interval is:
[tex]\[ \boxed{x \leq -3 \quad \text{or} \quad 2 \leq x < 7} \][/tex]
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