What is x<sqrt(2-x) ?

The solution to x<sqrt(2-x) is x<1

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x<sqrt(2-x)

x < 2 - x

x < 1

Interval Notation

(- \infty, 1)

Solution steps

x < 2 - x

Switch sides

2 - x > x

f (x) > g (x)

then

g (x) < 0 or g (x) \geq 0

For

x < 0 : x < 0

For

x \geq 0 : 0 \leq x < 1

Find singularity points

Find non-negative values for radicals:

x \leq 2

Combine the intervals

(0 \leq x < 1 or x < 0) and x \leq 2

Merge Overlapping Intervals

x < 1

Verify Solutions:

x < 1

True

The solution is

x < 1

- 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 )}

x - 3 < (15 - x)

x - 2 < x - 4