Question

3/x+2>0???????????????????​

Answer 1

$\dfrac3{x+2}>0\;\Rightarrow\;x>-2\\\\x\in(-2;\;+\infty)$

Answer 2

It looks like you've written an inequality: $\frac{3}{x+2} > 0$. To solve this inequality for $x$, you can follow these steps:

1. Find the critical points: Determine the values of $x$ that make the denominator ($x+2$) equal to zero. In this case, $x = -2$.

2. Test intervals: Divide the number line into intervals using the critical points you found. Test a value from each interval in the inequality to determine whether it's true or false.

3. Analyze signs: Based on the results from step 2, determine when the expression $\frac{3}{x+2}$ is positive (greater than zero) and when it's negative (less than zero) within each interval.

4. Final solution: Determine the intervals where the expression $\frac{3}{x+2}$ is greater than zero.

Let's go through it step by step:

1. Critical point: $x = -2$.

2. Test intervals:

   • Choose a value less than -2, e.g., $x = -3$: $\frac{3}{-3 + 2} = -3 < 0$.
   • Choose a value between -2 and a larger number, e.g., $x = 0$: $\frac{3}{0 + 2} = \frac{3}{2} > 0$.
   • Choose a value greater than -2, e.g., $x = 1$: $\frac{3}{1 + 2} = 1 > 0$.

3. Analyze signs:

   • $x < -2$: Negative
   • $-2 < x < 0$: Positive
   • $x > 0$: Positive

4. Final solution: The expression $\frac{3}{x+2}$ is greater than zero when $x$ is in the interval $-2 < x < 0$ or $x > 0$.

In interval notation, the solution is $x \in (-2, 0) \cup (0, \infty)$.

This means that the inequality $\frac{3}{x+2} > 0$ is satisfied when $x$ is any value greater than 0 or any value between -2 and 0 (excluding -2 itself).