What is sqrt(x^2-x-12)<x ?

The solution to sqrt(x^2-x-12) = 4

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sqrt(x^2-x-12)<x

x^{2} - x - 12 < x

x \geq 4

Interval Notation

[4, \infty)

Solution steps

x^{2} - x - 12 < x

f (x) < g (x)

then

g (x) > 0

x > 0

x^{2} - x - 12 < x : x > - 12

Find singularity points

Find non-negative values for radicals:

x \leq - 3 or x \geq 4

Combine the intervals

x > 0 and x > - 12 and (x \leq - 3 or x \geq 4)

Merge Overlapping Intervals

x \geq 4

Verify Solutions:

x \geq 4

True

The solution is

x \geq 4

15 \cdot 2

3 (2 u + 5) = 33

∣ 3 x + 9 ∣ < 12

(s^{2} + 1) (s^{3} + 1)

ln (1 + 0)