What is sqrt(x-2)+2<= x ?

The solution to sqrt(x-2)+2 = 3

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sqrt(x-2)+2<= x

x - 2 + 2 \leq x

x = 2 or x \geq 3

Interval Notation

x = 2 \cup [3, \infty)

Solution steps

x - 2 + 2 \leq x

Subtract

2

from both sides

x - 2 + 2 - 2 \leq x - 2

Simplify

x - 2 \leq x - 2

f (x) \leq g (x)

then

g (x) \geq 0

x - 2 \geq 0 : x \geq 2

x - 2 \leq x - 2 : x \leq 2 or x \geq 3

Find singularity points

Find non-negative values for radicals:

x \geq 2

Combine the intervals

x \geq 2 and (x \leq 2 or x \geq 3) and x \geq 2

Merge Overlapping Intervals

x = 2 or x \geq 3

Verify Solutions:

x = 2

True

, x \geq 3

True

The solution is

x = 2 or x \geq 3

x - 3 < x - 2

x^{2} + 1 > x - 1

x^{2} + 1 \geq 0

x < 5

5 x - 4 + 3 x + 1 < 3