Answer :
Sure, let’s solve the inequality [tex]\(\frac{x^2+7x+10}{x+\frac{2}{3}}>0\)[/tex] step-by-step.
### Step 1: Find the Roots of the Numerator and the Denominator
First, we identify the roots of both the numerator [tex]\(x^2 + 7x + 10\)[/tex] and the denominator [tex]\(x + \frac{2}{3}\)[/tex].
Numerator:
[tex]\[ x^2 + 7x + 10 = 0 \][/tex]
This is a quadratic equation. We can factorize or use the quadratic formula:
[tex]\[ x^2 + 7x + 10 = (x + 2)(x + 5) = 0 \][/tex]
Setting each factor to zero gives:
[tex]\[ x + 2 = 0 \quad \text{or} \quad x + 5 = 0 \][/tex]
So the roots of the numerator are:
[tex]\[ x = -2 \quad \text{and} \quad x = -5 \][/tex]
Denominator:
[tex]\[ x + \frac{2}{3} = 0 \][/tex]
Solving for [tex]\(x\)[/tex]:
[tex]\[ x = -\frac{2}{3} \][/tex]
### Step 2: Determine the Critical Points and Intervals
The critical points are the roots of the numerator and denominator:
[tex]\[ x = -5, \, -2, \, -\frac{2}{3} \][/tex]
These points divide the number line into intervals. Let's determine these intervals:
[tex]\[ (-\infty, -5), \, (-5, -2), \, (-2, -\frac{2}{3}), \, (-\frac{2}{3}, \infty) \][/tex]
### Step 3: Test Each Interval
We need to determine the sign of the expression [tex]\(\frac{x^2+7x+10}{x+\frac{2}{3}}\)[/tex] in each interval. We do this by picking test points in each interval and substituting them into the inequality.
#### Interval [tex]\((-∞, -5)\)[/tex]:
Choose a test point: [tex]\(x = -6\)[/tex]
[tex]\[ \frac{(-6)^2 + 7(-6) + 10}{-6 + \frac{2}{3}} = \frac{36 - 42 + 10}{-6.67} = \frac{4}{-6.67} < 0 \][/tex]
#### Interval [tex]\((-5, -2)\)[/tex]:
Choose a test point: [tex]\(x = -3.5\)[/tex]
[tex]\[ \frac{(-3.5)^2 + 7(-3.5) + 10}{-3.5 + \frac{2}{3}} = \frac{12.25 - 24.5 + 10}{-2.83} = \frac{-2.25}{-2.83} > 0 \][/tex]
#### Interval [tex]\((-2, -\frac{2}{3})\)[/tex]:
Choose a test point: [tex]\(x = -1\)[/tex]
[tex]\[ \frac{(-1)^2 + 7(-1) + 10}{-1 + \frac{2}{3}} = \frac{1 - 7 + 10}{-\frac{1}{3}} = \frac{4}{-\frac{1}{3}} < 0 \][/tex]
#### Interval [tex]\((-\frac{2}{3}, ∞)\)[/tex]:
Choose a test point: [tex]\(x = 0\)[/tex]
[tex]\[ \frac{(0)^2 + 7(0) + 10}{0 + \frac{2}{3}} = \frac{10}{\frac{2}{3}} > 0 \][/tex]
### Step 4: Combine the Results
Based on the signs of the expression, we see:
- The inequality is negative on [tex]\((-∞, -5)\)[/tex] and [tex]\((-2, -\frac{2}{3})\)[/tex]
- The inequality is positive on [tex]\((-5, -2)\)[/tex] and [tex]\((-\frac{2}{3}, ∞)\)[/tex]
### Step 5: Consider the Boundaries
The expression [tex]\(\frac{x^2+7x+10}{x+\frac{2}{3}}\)[/tex] is undefined at [tex]\(x = -\frac{2}{3}\)[/tex].
The inequality [tex]\(\frac{x^2 + 7x + 10}{x + \frac{2}{3}} > 0\)[/tex] holds true where the expression is positive:
[tex]\[ x \in (-5, -2) \cup (-\frac{2}{3}, \infty) \][/tex]
Thus, the solution to the inequality [tex]\(\frac{x^2+7x+10}{x+\frac{2}{3}}>0\)[/tex] is:
[tex]\[ x \in (-5, -2) \cup \left(-\frac{2}{3}, +\infty\right) \][/tex]
### Step 1: Find the Roots of the Numerator and the Denominator
First, we identify the roots of both the numerator [tex]\(x^2 + 7x + 10\)[/tex] and the denominator [tex]\(x + \frac{2}{3}\)[/tex].
Numerator:
[tex]\[ x^2 + 7x + 10 = 0 \][/tex]
This is a quadratic equation. We can factorize or use the quadratic formula:
[tex]\[ x^2 + 7x + 10 = (x + 2)(x + 5) = 0 \][/tex]
Setting each factor to zero gives:
[tex]\[ x + 2 = 0 \quad \text{or} \quad x + 5 = 0 \][/tex]
So the roots of the numerator are:
[tex]\[ x = -2 \quad \text{and} \quad x = -5 \][/tex]
Denominator:
[tex]\[ x + \frac{2}{3} = 0 \][/tex]
Solving for [tex]\(x\)[/tex]:
[tex]\[ x = -\frac{2}{3} \][/tex]
### Step 2: Determine the Critical Points and Intervals
The critical points are the roots of the numerator and denominator:
[tex]\[ x = -5, \, -2, \, -\frac{2}{3} \][/tex]
These points divide the number line into intervals. Let's determine these intervals:
[tex]\[ (-\infty, -5), \, (-5, -2), \, (-2, -\frac{2}{3}), \, (-\frac{2}{3}, \infty) \][/tex]
### Step 3: Test Each Interval
We need to determine the sign of the expression [tex]\(\frac{x^2+7x+10}{x+\frac{2}{3}}\)[/tex] in each interval. We do this by picking test points in each interval and substituting them into the inequality.
#### Interval [tex]\((-∞, -5)\)[/tex]:
Choose a test point: [tex]\(x = -6\)[/tex]
[tex]\[ \frac{(-6)^2 + 7(-6) + 10}{-6 + \frac{2}{3}} = \frac{36 - 42 + 10}{-6.67} = \frac{4}{-6.67} < 0 \][/tex]
#### Interval [tex]\((-5, -2)\)[/tex]:
Choose a test point: [tex]\(x = -3.5\)[/tex]
[tex]\[ \frac{(-3.5)^2 + 7(-3.5) + 10}{-3.5 + \frac{2}{3}} = \frac{12.25 - 24.5 + 10}{-2.83} = \frac{-2.25}{-2.83} > 0 \][/tex]
#### Interval [tex]\((-2, -\frac{2}{3})\)[/tex]:
Choose a test point: [tex]\(x = -1\)[/tex]
[tex]\[ \frac{(-1)^2 + 7(-1) + 10}{-1 + \frac{2}{3}} = \frac{1 - 7 + 10}{-\frac{1}{3}} = \frac{4}{-\frac{1}{3}} < 0 \][/tex]
#### Interval [tex]\((-\frac{2}{3}, ∞)\)[/tex]:
Choose a test point: [tex]\(x = 0\)[/tex]
[tex]\[ \frac{(0)^2 + 7(0) + 10}{0 + \frac{2}{3}} = \frac{10}{\frac{2}{3}} > 0 \][/tex]
### Step 4: Combine the Results
Based on the signs of the expression, we see:
- The inequality is negative on [tex]\((-∞, -5)\)[/tex] and [tex]\((-2, -\frac{2}{3})\)[/tex]
- The inequality is positive on [tex]\((-5, -2)\)[/tex] and [tex]\((-\frac{2}{3}, ∞)\)[/tex]
### Step 5: Consider the Boundaries
The expression [tex]\(\frac{x^2+7x+10}{x+\frac{2}{3}}\)[/tex] is undefined at [tex]\(x = -\frac{2}{3}\)[/tex].
The inequality [tex]\(\frac{x^2 + 7x + 10}{x + \frac{2}{3}} > 0\)[/tex] holds true where the expression is positive:
[tex]\[ x \in (-5, -2) \cup (-\frac{2}{3}, \infty) \][/tex]
Thus, the solution to the inequality [tex]\(\frac{x^2+7x+10}{x+\frac{2}{3}}>0\)[/tex] is:
[tex]\[ x \in (-5, -2) \cup \left(-\frac{2}{3}, +\infty\right) \][/tex]
