What is sqrt(x+3)+4>= 5 ?

The solution to sqrt(x+3)+4>= 5 is x>=-2

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sqrt(x+3)+4>= 5

x + 3 + 4 \geq 5

x \geq - 2

Interval Notation

[- 2, \infty)

Solution steps

x + 3 + 4 \geq 5

Move

4

to the right side

x + 3 \geq 1

Square both sides

(x + 3)^{2} \geq 1^{2}

Simplify

x + 3 \geq 1

Move

3

to the right side

x \geq - 2

Find singularity points

Find non-negative values for radicals:

x \geq - 3

Combine the intervals

x \geq - 2 and x \geq - 3

Merge Overlapping Intervals

x \geq - 2

0. 3^{- 0.35}

2 \leq x + 5 < 9

x < 4 or x < - 6

(2 x - 16) + (4 x + 4) = 180

(x + 1)^{2} = 13 - 2 x