What is |x^2-1|+|x^2-16|>= 2 ?

The solution to |x^2-1|+|x^2-16|>= 2 is True for all x

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|x^2-1|+|x^2-16|>= 2

x^{2} - 1 + x^{2} - 16 \geq 2

True for all x

Interval Notation

(- \infty, \infty)

Solution steps

x^{2} - 1 + x^{2} - 16 \geq 2

Find positive and negative intervals

x \leq - 4, - 4 < x \leq - 1, - 1 < x < 1, 1 \leq x < 4, x \geq 4

Solve the inequality for each interval

x \leq - 4 or - 4 < x \leq - 1 or - 1 < x < 1 or 1 \leq x < 4 or x \geq 4

Merge Overlapping Intervals

True for all x

- 3 w - 4 = w + 3

(4 + 2 i) + (- 3 - 5 i)

\frac{1}{2} - \frac{1}{3} + \frac{1}{4} - \frac{1}{5} + \frac{1}{6}

1 + t + 9 + 10 t

What is |x^2-1|+|x^2-16|>= 2 ?
The solution to |x^2-1|+|x^2-16|>= 2 is True for all x