What is 1/(sqrt(x^2-1))+3>= 0 ?

The solution to 1/(sqrt(x^2-1))+3>= 0 is x = 1

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1/(sqrt(x^2-1))+3>= 0

\frac{1}{x ^{2} - 1} + 3 \geq 0

x \leq - 1 or x \geq 1

Interval Notation

(- \infty, - 1] \cup [1, \infty)

Solution steps

\frac{1}{x ^{2} - 1} + 3 \geq 0

Move

3

to the right side

\frac{1}{x ^{2} - 1} \geq - 3

\frac{1}{a} \geq 0

then

a > 0

x^{2} - 1 > - 3

If n is even,

for all

u

True for all x

Find singularity points

Find non-negative values for radicals:

x \leq - 1 or x \geq 1

Combine the intervals

True for all x and (x \leq - 1 or x \geq 1)

Merge Overlapping Intervals

x \leq - 1 or x \geq 1

Find singularity points

Find non-negative values for radicals:

x \leq - 1 or x \geq 1

Combine the intervals

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

Merge Overlapping Intervals

x \leq - 1 or x \geq 1

x^{- \frac{2}{3}}

16 x^{3} - 16 x^{2} - 25 x + 25

16 x^{3} - 100 x

8 s = 48

\frac{m}{6} = \frac{m + 1}{5}

What is 1/(sqrt(x^2-1))+3>= 0 ?
The solution to 1/(sqrt(x^2-1))+3>= 0 is x<=-1\lor x>= 1