What is sqrt(x+2)<sqrt(x+5) ?

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sqrt(x+2)<sqrt(x+5)

x + 2 < x + 5

x \geq - 2

Interval Notation

Solution steps

x + 2 < x + 5

Square both sides

(x + 2)^{2} < (x + 5)^{2}

Simplify

x + 2 < x + 5

x + 2 < x + 5 :

True for all

x

True for all x

Find singularity points

Find non-negative values for radicals:

x \geq - 2

Combine the intervals

True for all x and x \geq - 2

Merge Overlapping Intervals

x \geq - 2

2 - x > x

\frac{( 2 x - 1 )}{( x - 2 )} < 1

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

8 - 3 x < - 9

(x + 2) \leq x + 2