What is sqrt(4-x^2)>= 0 ?

The solution to sqrt(4-x^2)>= 0 is -2<= x<= 2

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sqrt(4-x^2)>= 0

4 - x^{2} \geq 0

- 2 \leq x \leq 2

Interval Notation

[- 2, 2]

Solution steps

4 - x^{2} \geq 0

If n is even,

for all

u

True for all x

Find singularity points

Find non-negative values for radicals:

- 2 \leq x \leq 2

Combine the intervals

True for all x and - 2 \leq x \leq 2

Merge Overlapping Intervals

- 2 \leq x \leq 2

\frac{1}{3} \leq \frac{x + 1}{5} \leq \frac{5}{4}

\frac{π}{5} + \frac{π}{10}

(\frac{x ^{6} y ^{(- 3)}}{8 y ^{\frac{3}{2}}})^{- \frac{1}{3}}

c^{2} + 9 c + 10 = - 2

3 x - 4 = x + 5