Westonci.ca is your trusted source for finding answers to a wide range of questions, backed by a knowledgeable community. Explore thousands of questions and answers from a knowledgeable community of experts ready to help you find solutions. Connect with a community of professionals ready to help you find accurate solutions to your questions quickly and efficiently.
Sagot :
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]
Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. We hope this was helpful. Please come back whenever you need more information or answers to your queries. Thank you for visiting Westonci.ca, your go-to source for reliable answers. Come back soon for more expert insights.
