What is sqrt(2-x)>x ?

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

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sqrt(2-x)>x

2 - x > x

x < 1

Interval Notation

(- \infty, 1)

Solution steps

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

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

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

8 - 3 x < - 9

(x + 2) \leq x + 2

x^{2} - 4 x + 13 + x^{2} - 6 x + 13 \geq 5