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# 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

x^{2 }> 2

⇒ x < -√2 or x > √2

Hence, x ∈ (-∞, -√2) ∪ (√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 (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.

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