If `x=-3` when `x^(2)+2xr+r^(2)=0`, what is the value of r?

Explanation

To solve the equation \( x^2 + 2xr + r^2 = 0 \) for the value of \( r \) when \( x = -3 \), follow these steps:

### Step 1: Substitute the value of \( x \)
Substituting \( x = -3 \) into the equation gives:
\[ (-3)^2 + 2(-3)r + r^2 = 0 \]
Calculating \( (-3)^2 \):
\[ 9 + 2(-3)r + r^2 = 0 \]

### Step 2: Simplify the equation
Now simplify the equation:
\[ 9 - 6r + r^2 = 0 \]
Rearranging gives:
\[ r^2 - 6r + 9 = 0 \]

### Step 3: Factor the quadratic equation
The equation \( r^2 - 6r + 9 = 0 \) can be factored as:
\[ (r - 3)(r - 3) = 0 \]
This implies:
\[ (r - 3)^2 = 0 \]

### Step 4: Solve for \( r \)
Setting the factor equal to zero gives:
\[ r - 3 = 0 \implies r = 3 \]

### Final Answer
The value of \( r \) is:
\[ \boxed{3} \]

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