What is sqrt(x-3)<(15-x) ?

The solution to sqrt(x-3)<(15-x) is 3<= x<12

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sqrt(x-3)<(15-x)

x - 3 < (15 - x)

3 \leq x < 12

Interval Notation

[3, 12)

Solution steps

x - 3 < 15 - x

f (x) < g (x)

then

g (x) > 0

15 - x > 0 : x < 15

x - 3 < 15 - x : x < 12 or x > 19

Find singularity points

Find non-negative values for radicals:

x \geq 3

Combine the intervals

x < 15 and (x < 12 or x > 19) and x \geq 3

Merge Overlapping Intervals

3 \leq x < 12

Verify Solutions:

3 \leq x < 12

True

The solution is

3 \leq x < 12

x - 2 < x - 4

x^{2} - 3 x + 2 > 2 x - 5

x - 4 < 2

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

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