What is (x+2)<= sqrt(x+2) ?

The solution to (x+2)<= sqrt(x+2) is -2<= x<=-1

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(x+2)<= sqrt(x+2)

(x + 2) \leq x + 2

- 2 \leq x \leq - 1

Interval Notation

[- 2, - 1]

Solution steps

x + 2 \leq x + 2

Switch sides

x + 2 \geq x + 2

f (x) \geq g (x)

then

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

For

x + 2 < 0 : x < - 2

For

x + 2 \geq 0 : - 2 \leq x \leq - 1

Find singularity points

Find non-negative values for radicals:

x \geq - 2

Combine the intervals

(- 2 \leq x \leq - 1 or x < - 2) and x \geq - 2

Merge Overlapping Intervals

- 2 \leq x \leq - 1

Verify Solutions:

- 2 \leq x \leq - 1

True

The solution is

- 2 \leq x \leq - 1

x^{2} - 4 x + 13 + x^{2} - 6 x + 13 \geq 5

(4 - x) (x - 3)^{\frac{1}{2}} < 0

\frac{( x - 1 )}{( x + 1 )} > 0

x - 1 > 3 - x

x^{2} - 5 x + 4 \leq 2 x - 2