What is |x^2+x-2|<= |x^2|+|x+2| ?

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|x^2+x-2|<= |x^2|+|x+2|

x^{2} + x - 2 \leq x^{2} + ∣ x + 2 ∣

x \leq - 1 or x \geq 0

Interval Notation

(- \infty, - 1] \cup [0, \infty)

Solution steps

x^{2} + x - 2 \leq x^{2} + ∣ x + 2 ∣

Find positive and negative intervals

x \leq - 2, - 2 < x < 1, x \geq 1

Solve the inequality for each interval

x \leq - 2 or (- 2 < x \leq - 1 or 0 \leq x < 1) or x \geq 1

Merge Overlapping Intervals

x \leq - 1 or x \geq 0

8^{2} + x^{2} = 1 0^{2}

(x + 1)^{2} - 4

1 6^{l o g_{4} (8)}

8 b^{3} c^{2} + 4 b^{3} c^{2}

- 4 < 2 - 2 x \leq 10

What is |x^2+x-2|<= |x^2|+|x+2| ?
The solution to |x^2+x-2|<= |x^2|+|x+2| is x<=-1\lor x>= 0