If R is the set of real numbers, then which elements are presented by theset `{x in R : x^(2) + 2x +2 =0}`

Answer

To solve the equation \( x^2 + 2x + 2 = 0 \) and determine the elements of the set \( \{ x \in \mathbb{R} : x^2 + 2x + 2 = 0 \} \), we can follow these steps:

### Step 1: Rewrite the equation
First, we rewrite the quadratic equation:
\[ x^2 + 2x + 2 = 0 \]

### Step 2: Complete the square
Next, we complete the square for the quadratic expression:
\[ x^2 + 2x + 1 + 1 = 0 \]
This can be rewritten as:
\[ (x + 1)^2 + 1 = 0 \]

### Step 3: Analyze the equation
Now, we analyze the equation:
\[ (x + 1)^2 + 1 = 0 \]
Here, \( (x + 1)^2 \) is always non-negative (i.e., it is greater than or equal to 0) because it is a square term. The smallest value it can take is 0, which occurs when \( x + 1 = 0 \) or \( x = -1 \).

### Step 4: Determine the impossibility
Since \( (x + 1)^2 \geq 0 \), we have:
\[ (x + 1)^2 + 1 \geq 1 \]
This means that the left-hand side can never equal 0, as it is always at least 1. Therefore, there are no real solutions to the equation.

### Conclusion
Thus, the set of elements that satisfy the equation is the empty set:
\[ \{ x \in \mathbb{R} : x^2 + 2x + 2 = 0 \} = \emptyset \]

---