What is sqrt(x-4)<2 ?

The solution to sqrt(x-4)<2 is 4<= x<8

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sqrt(x-4)<2

x - 4 < 2

4 \leq x < 8

Interval Notation

[4, 8)

Solution steps

x - 4 < 2

Square both sides

(x - 4)^{2} < 2^{2}

Simplify

x - 4 < 4

Move

4

to the right side

x < 8

Find singularity points

Find non-negative values for radicals:

x \geq 4

Combine the intervals

x < 8 and x \geq 4

Merge Overlapping Intervals

4 \leq x < 8

(x^{2} - 1) \cdot x^{2} - x - 2 \geq 0

\frac{( 1 - 21 - 4 x - x ^{2} )}{( x + 1 )} \geq 0

x - 1 - ∣ x ∣ \leq 0

3 - x \leq 1 - x

x + 3 + x + 2 - 2 x + 4 > 0