What is sqrt(5-x^2)>x-1 ?

The solution to sqrt(5-x^2)>x-1 is -sqrt(5)<= x<2

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sqrt(5-x^2)>x-1

5 - x^{2} > x - 1

- 5 \leq x < 2

Interval Notation

[- 5, 2)

Decimal Notation

- 2.23606 \dots \leq x < 2

Solution steps

5 - x^{2} > x - 1

f (x) > g (x)

then

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

For

x - 1 < 0 : x < 1

For

x - 1 \geq 0 : 1 \leq x < 2

Find singularity points

Find non-negative values for radicals:

- 5 \leq x \leq 5

Combine the intervals

(1 \leq x < 2 or x < 1) and - 5 \leq x \leq 5

Merge Overlapping Intervals

- 5 \leq x < 2

Verify Solutions:

- 5 \leq x < 2

True

The solution is

- 5 \leq x < 2

\frac{- ( x + 6 )}{( x - 2 )} < 3

x + 4 x - 4 \geq 0

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

x \geq 4 - x

(3) x + 3 + x^{2} + 4 x - 5 > 0