The set of all real numbers x for which x² - |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, x² – |x + 2| + x > 0

We can discuss this problem by observing two different cases.

Case 1: When (x + 2) ≥ 0.

Therefore, x² – x – 2 + x > 0

Hence, x² – 2 > 0

x² > 2

⇒ x < -√2 or x > √2

Hence, x ∈ (-∞, -√2) ∪ (√2, ∞) ……. (1)

Case 2: When (x + 2) < 0

Then x² + x + 2 + x > 0

So, x² + 2x + 2 > 0

This gives (x + 1)² + 1 > 0 and this is true for every x (Because (x + 1)² ≥ 0 ⇒(x + 1)²  + 1 ≥ 1 > 0)

Hence, x ∈ (-∞, ∞) ……… (2)

From equations (1) and (2), we take the intersection, hence we get x ∈ (-∞, -√2) ∪ (√2, ∞).

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