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

Answer

To determine for how many integers \( n \) the expression \( \sqrt{9 - (n + 2)^2} \) is a real number, we need to ensure that the quantity under the square root is non-negative. This leads us to the following steps:

### Step 1: Set up the inequality
We start with the condition for the square root to be real:
\[ 9 - (n + 2)^2 \geq 0 \]

### Step 2: Rearrange the inequality
Rearranging gives us:
\[ (n + 2)^2 \leq 9 \]

### Step 3: Take the square root
Taking the square root of both sides (and remembering to consider both the positive and negative roots), we have:
\[ -n - 2 \leq 3 \quad \text{and} \quad n + 2 \leq 3 \]
\[ -n - 2 \geq -3 \quad \text{and} \quad n + 2 \geq -3 \]

### Step 4: Solve the inequalities
Now we solve these inequalities one by one.

1. From \( (n + 2)^2 \leq 9 \):
- \( n + 2 \leq 3 \) leads to:
\[ n \leq 1 \]
- \( n + 2 \geq -3 \) leads to:
\[ n \geq -5 \]

### Step 5: Combine the results
Combining these results, we find:
\[ -5 \leq n \leq 1 \]

### Step 6: Identify integer solutions
Now we need to find the integer values of \( n \) within this range:
- The integers that satisfy \( -5 \leq n \leq 1 \) are:
\[ -5, -4, -3, -2, -1, 0, 1 \]

### Step 7: Count the integers
Counting these integers gives us a total of:
\[ 7 \text{ integers} \]

### Conclusion
Thus, the total number of integers \( n \) for which \( \sqrt{9 - (n + 2)^2} \) is a real number is \( \boxed{7} \).

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