What is x^2-|x+2|+x=>0 ?

The solution to x^2-|x+2|+x=>0 is x sqrt(2)

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x^2-|x+2|+x=>0

x^{2} - ∣ x + 2 ∣ + x => 0

x < - 2 or x > 2

Interval Notation

(- \infty, - 2) \cup (2, \infty)

Decimal Notation

x < - 1.41421 \dots or x > 1.41421 \dots

Solution steps

x^{2} - ∣ x + 2 ∣ + x > 0

Find positive and negative intervals

x < - 2, x \geq - 2

Solve the inequality for each interval

x < - 2 or (- 2 \leq x < - 2 or x > 2)

Merge Overlapping Intervals

x < - 2 or x > 2

∣ 4 - x ∣ < - 11

\frac{( x - 2 )}{( x - 2 )} \cdot (\frac{- 1}{2}) \leq 0

x^{2} - 4 \leq 0

(∣ x ∣ - 5) (∣ x ∣ - 7) \leq 0

(c) ∣ x - 3 ∣ - ∣ x + 2 ∣ \leq 3

What is x^2-|x+2|+x=>0 ?
The solution to x^2-|x+2|+x=>0 is x<-sqrt(2)\lor x>sqrt(2)