What is x^2+x>|x+2| ?

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

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x^2+x>|x+2|

x^{2} + x > ∣ x + 2 ∣

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 > ∣ x + 2 ∣

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

∣ r ∣ < 1

\frac{2}{∣ x + 2 ∣} \leq 1

∣ x + 6 ∣ + ∣ x - 1 ∣ \leq 2 x + 5

x^{2} - ∣ 5 x - 3 ∣ - x < 2

\frac{( ∣ x + 4 ∣ + 2 x )}{x + 1} > 2