What is 3sqrt(a)>= 12 ?

The solution to 3sqrt(a)>= 12 is a>= 16

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3sqrt(a)>= 12

3 a \geq 12

a \geq 16

Interval Notation

[16, \infty)

Solution steps

3 a \geq 12

Divide both sides by

3

\frac{3 a}{3} \geq \frac{12}{3}

Simplify

a \geq 4

Square both sides

(a)^{2} \geq 4^{2}

Simplify

a \geq 16

Find singularity points

Find non-negative values for radicals:

a \geq 0

Combine the intervals

a \geq 16 and a \geq 0

Merge Overlapping Intervals

a \geq 16

(x + 2)^{2} - 2

\frac{3}{4} z + 6 \geq - \frac{3}{4} z

5 x \geq 7 x - 4

- 16 t^{2} + 70 t + 5 = 30

3 x + 2 < - 7