What is sqrt(x^2-x-16)>2 ?

The solution to sqrt(x^2-x-16)>2 is x 5

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sqrt(x^2-x-16)>2

x^{2} - x - 16 > 2

x < - 4 or x > 5

Solution steps

x^{2} - x - 16 > 2

Square both sides

(x^{2} - x - 16)^{2} > 2^{2}

Simplify

x^{2} - x - 16 > 4

x^{2} - x - 16 > 4 : x < - 4 or x > 5

x < - 4 or x > 5

Find singularity points

Find non-negative values for radicals:

x \leq \frac{- 65 + 1}{2} or x \geq \frac{65 + 1}{2}

Combine the intervals

(x < - 4 or x > 5) and (x \leq \frac{- 65 + 1}{2} or x \geq \frac{65 + 1}{2})

Merge Overlapping Intervals

x < - 4 or x > 5

3 x - 2 + x = 4 - 7 x

3 = 2 x + 3

3 (4 x + 7) + 1 = 8 x + 24

12 = 8 k

7 - y = 3