For how many integers n is `sqrt(9-(n+2)^2)` a real number?

Answer

To find how many integers \( n \) make \( \sqrt{9 - (n + 2)^2} \) a real number, we need to ensure that the expression inside the square root is non-negative. Thus, we need to solve the inequality:

\[ 9 - (n + 2)^2 \geq 0 \]

### Step 1: Rearranging the Inequality
First, we can rearrange the inequality:

\[ (n + 2)^2 \leq 9 \]

### Step 2: Taking Square Roots
Next, we take the square root of both sides. Remember that when we take the square root, we consider both the positive and negative roots:

\[ -n - 2 \leq 3 \quad \text{and} \quad n + 2 \leq 3 \]

This gives us two inequalities to solve.

### Step 3: Solving the First Inequality
For the first inequality:

\[ n + 2 \leq 3 \]

Subtracting 2 from both sides gives:

\[ n \leq 1 \]

### Step 4: Solving the Second Inequality
Now, for the second inequality:

\[ -n - 2 \geq -3 \]

Adding 2 to both sides gives:

\[ -n \geq -1 \]

Multiplying both sides by -1 (which reverses the inequality) gives:

\[ n \leq 1 \]

### Step 5: Solving the Combined Inequalities
Now we combine the inequalities we found:

\[ -3 \leq n \leq 1 \]

### Step 6: Finding Integer Solutions
The integers that satisfy this inequality are:

\[ -3, -2, -1, 0, 1 \]

### Step 7: Counting the Integer Solutions
Counting these integers gives us:

\[ -3, -2, -1, 0, 1 \quad \text{(5 integers)} \]

Thus, the total number of integers \( n \) for which \( \sqrt{9 - (n + 2)^2} \) is a real number is **5**.

### Summary of Steps
1. Set up the inequality \( 9 - (n + 2)^2 \geq 0 \).
2. Rearrange to \( (n + 2)^2 \leq 9 \).
3. Solve \( n + 2 \leq 3 \) and \( n + 2 \geq -3 \).
4. Combine the results to find the range of \( n \).
5. Identify the integers within that range.