What is sqrt(2-x)<x ?

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

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sqrt(2-x)<x

2 - x < x

1 < x \leq 2

Interval Notation

(1, 2]

Solution steps

2 - x < x

f (x) < g (x)

then

g (x) > 0

x > 0

2 - x < x : x < - 2 or x > 1

Find singularity points

Find non-negative values for radicals:

x \leq 2

Combine the intervals

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

Merge Overlapping Intervals

1 < x \leq 2

Verify Solutions:

1 < x \leq 2

True

The solution is

1 < x \leq 2

4 t = 12 - 4 t

10 x^{2} - 3 x - 27

n^{2} - 100

2 x^{2} - 4 x = - 1

- (x - (1 0^{- 5}))^{2} = (3.12 \cdot 1 0^{- 5})