# The set of all real numbers x for which x^{2} - |x + 2| + x > 0 is

# 1). (-∞, -2) ∪ (2, ∞)

# 2. (-∞, -√2) ∪ (√2, ∞)

# 3. (-∞, -1) ∪ (1, ∞)

# 4. (√2, ∞)

Real numbers are the combination of rational numbers as well as irrational numbers.

## Answer: Option (2) , x ∈ (-∞, -√2) ∪ (√2, ∞)

Let us proceed step by step

**Explanation:**

From the given data, x^{2} – |x + 2| + x > 0

We can discuss this problem by observing two different cases.

Case 1: When (x + 2) ≥ 0.

Therefore, x^{2} – x – 2 + x > 0

Hence, x^{2} – 2 > 0

So, either x = √2 or x = - √2

Hence, x ∈ (-√2, 0) ∪ (√2, ∞) ……. (1)

Case 2: When (x + 2) < 0

Then x^{2} + x + 2 + x > 0

So, x^{2} + 2x + 2 > 0

This gives (x + 1)^{2} + 1 > 0 and this is true for every x

Hence, x ≤ -2 or x ∈ (-∞, -2) ……… (2)

From equations (2) and (3) we get x ∈ (-∞, -√2) ∪ (√2, ∞).

### Therefore, x ∈ (-∞, -√2) ∪ (√2, ∞) is the required answer.