What is 1/(sqrt(x-2)-4)<= 0 ?

The solution to 1/(sqrt(x-2)-4)<= 0 is 2<= x<18

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1/(sqrt(x-2)-4)<= 0

\frac{1}{x - 2 - 4} \leq 0

2 \leq x < 18

Solution steps

\frac{1}{x - 2 - 4} \leq 0

Let

u = x - 2

\frac{1}{u - 4} \leq 0

\frac{1}{u - 4} \leq 0 : u < 4

u < 4

Substitute back

u = x - 2

x - 2 < 4

Square both sides

(x - 2)^{2} < 4^{2}

Simplify

x - 2 < 16

Move

2

to the right side

x < 18

Find singularity points

Find non-negative values for radicals:

x \geq 2

Combine the intervals

x < 18 and x \geq 2

Merge Overlapping Intervals

2 \leq x < 18

- \frac{1}{2} \cdot \frac{3}{5}

\frac{7 + 5 i}{1 + i}

0.125 x = 0.025 (5 x + 1)

\frac{7}{18} + \frac{7}{36}

- 2 x \leq - 6 or - 2 x \geq 0