What is sqrt(2+x-x^2)<x+2 ?

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

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sqrt(2+x-x^2)<x+2

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

- 1 \leq x \leq 2

Interval Notation

[- 1, 2]

Solution steps

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

f (x) < g (x)

then

g (x) > 0

x + 2 > 0 : x > - 2

2 + x - x^{2} < x + 2 :

True for all

x

Find singularity points

Find non-negative values for radicals:

- 1 \leq x \leq 2

Combine the intervals

x > - 2 and True for all x and - 1 \leq x \leq 2

Merge Overlapping Intervals

- 1 \leq x \leq 2

Verify Solutions:

- 1 \leq x \leq 2

True

The solution is

- 1 \leq x \leq 2

2 x + 14 > x + 3

\frac{( x - 3 )}{( x - 2 )} > 1

x > (2 - x)^{\frac{1}{2}}

x - 1 - 8 - 2 x \geq 1

\frac{2}{( 3 x - 5 )} \geq 0