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

The solution to sqrt(x)<sqrt(2) is 0<= x<2

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sqrt(x)<sqrt(2)

x < 2

0 \leq x < 2

Interval Notation

[0, 2)

Solution steps

x < 2

Square both sides

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

Simplify

x < (2)^{2}

Expand

(2)^{2} : 2

x < 2

Find singularity points

Find non-negative values for radicals:

x \geq 0

Combine the intervals

x < 2 and x \geq 0

Merge Overlapping Intervals

0 \leq x < 2

\frac{x}{2} > 4

x < 2 - x

- x^{2} + 8 x - 7 < x - 1

\frac{( ∣ x + 2 ∣ - ∣ x ∣ )}{( 8 - x ^{2} )} \geq 0

\frac{( x - 3 )}{( x + 1 )} \geq \frac{( x + 2 )}{( x - 4 )}