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

∣ x - 1 ∣ - ∣ x - 2 ∣ + ∣ x - 3 ∣ \geq 6

x^{2} - x \leq x + 1

∣ 2 x - 3 ∣ > ∣ x + 4 ∣

∣ - 0.5 x ∣ < 1

[\frac{( x ^{2} - 5 x + 6 )}{( ∣ x ∣ + 7 )}] < 0

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)