What is sqrt(x^2-2x-15)>2x+5 ?

The solution to sqrt(x^2-2x-15)>2x+5 is x<=-3

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sqrt(x^2-2x-15)>2x+5

x^{2} - 2 x - 15 > 2 x + 5

x \leq - 3

Interval Notation

(- \infty, - 3]

Solution steps

x^{2} - 2 x - 15 > 2 x + 5

f (x) > g (x)

then

g (x) < 0 or g (x) \geq 0

For

2 x + 5 < 0 : x < - \frac{5}{2}

For

2 x + 5 \geq 0 :

No Solution

Find singularity points

Find non-negative values for radicals:

x \leq - 3 or x \geq 5

Combine the intervals

(No Solution or x < - \frac{5}{2}) and (x \leq - 3 or x \geq 5)

Merge Overlapping Intervals

x \leq - 3

Verify Solutions:

x \leq - 3

True

The solution is

x \leq - 3

4 (1 - 6 n) + 6

49 x^{2} + 84 x y + 36 y^{2}

∣ 5 u + 10 ∣ > 15

x^{2} + 6 x - 27 < 0

- 1 < 1 - p