What is sqrt(8x-16)>= 4 ?

The solution to sqrt(8x-16)>= 4 is x>= 4

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sqrt(8x-16)>= 4

8 x - 16 \geq 4

x \geq 4

Interval Notation

[4, \infty)

Solution steps

8 x - 16 \geq 4

Square both sides

(8 x - 16)^{2} \geq 4^{2}

Simplify

8 x - 16 \geq 16

8 x - 16 \geq 16 : x \geq 4

x \geq 4

Find singularity points

Find non-negative values for radicals:

x \geq 2

Combine the intervals

x \geq 4 and x \geq 2

Merge Overlapping Intervals

x \geq 4

\frac{5}{4} - \frac{1}{4}

\frac{3 π}{4} + 4 π

16 (t - 3) = - 16

90 \cdot y = 450

\frac{1}{8} \cdot 4