What is (1+x)^2>= |1-x^2| ?

The solution to (1+x)^2>= |1-x^2| is x=-1\lor x>= 0

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(1+x)^2>= |1-x^2|

(1 + x)^{2} \geq 1 - x^{2}

x = - 1 or x \geq 0

Interval Notation

x = - 1 \cup [0, \infty)

Solution steps

(1 + x)^{2} \geq 1 - x^{2}

Find positive and negative intervals

x < - 1, - 1 \leq x \leq 1, x > 1

Solve the inequality for each interval

No Solution or (x = - 1 or 0 \leq x \leq 1) or x > 1

Merge Overlapping Intervals

x = - 1 or x \geq 0

x^{2} - 6 x + 8 \leq 4 - x

x^{3} + 1 < x^{2} - x + 1

∣ x - 1 ∣ + ∣ x + 2 ∣ \geq 3

\frac{1}{( 2 - ∣ x ∣ )} \geq 1

x^{2} - 3 x - 3 > x^{2} + 7 x - 13