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

The solution to x+4sqrt(x-4)>= 0 is x>= 4

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x+4sqrt(x-4)>= 0

x + 4 x - 4 \geq 0

x \geq 4

Interval Notation

[4, \infty)

Solution steps

x + 4 x - 4 \geq 0

Subtract

x

from both sides

x + 4 x - 4 - x \geq 0 - x

Simplify

4 x - 4 \geq - x

Divide both sides by

4

\frac{4 x - 4}{4} \geq \frac{- x}{4}

Simplify

x - 4 \geq - \frac{x}{4}

f (x) \geq g (x)

then

g (x) < 0 or g (x) \geq 0

For

- \frac{x}{4} < 0 : x > 0

For

- \frac{x}{4} \geq 0 :

No Solution

Find singularity points

Find non-negative values for radicals:

x \geq 4

Combine the intervals

(No Solution or x > 0) and x \geq 4

Merge Overlapping Intervals

x \geq 4

Verify Solutions:

x \geq 4

True

The solution is

x \geq 4

x^{2} + 3 x + 3 < 2 x + 1

x \geq 4 - x

(3) x + 3 + x^{2} + 4 x - 5 > 0

x + 2 < x + 5

2 - x > x