What is sqrt(x+2)>x ?

The solution to sqrt(x+2)>x is -2<= x<2

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sqrt(x+2)>x

x + 2 > x

- 2 \leq x < 2

Interval Notation

[- 2, 2)

Solution steps

x + 2 > x

f (x) > g (x)

then

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

For

x < 0 : x < 0

For

x \geq 0 : 0 \leq x < 2

Find singularity points

Find non-negative values for radicals:

x \geq - 2

Combine the intervals

(0 \leq x < 2 or x < 0) and x \geq - 2

Merge Overlapping Intervals

- 2 \leq x < 2

Verify Solutions:

- 2 \leq x < 2

True

The solution is

- 2 \leq x < 2

3 x^{2} + 8 x + 5

\frac{4 x + 5}{x - 8} \leq 0

3 i \cdot 4 i

x^{2} + x = - 9 x - 16

12 x^{2} - 228 x + 740 = 0