What is sqrt(4-x^2)>x+7 ?

The solution to sqrt(4-x^2)>x+7 is No Solution for x\in\mathbb{R}

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sqrt(4-x^2)>x+7

4 - x^{2} > x + 7

No Solution for x \in R

Solution steps

4 - x^{2} > x + 7

f (x) > g (x)

then

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

For

x + 7 < 0 : x < - 7

For

x + 7 \geq 0 :

No Solution

Find singularity points

Find non-negative values for radicals:

- 2 \leq x \leq 2

Combine the intervals

(No Solution or x < - 7) and - 2 \leq x \leq 2

Merge Overlapping Intervals

No Solution for x \in R

0.6 x^{2} - 1.5 x + 1 = 0

\frac{3 + 2}{1 - 3}

\frac{3 a}{8} - 2 = \frac{2 a + 9}{5}

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

x^{2} + \frac{3}{2} x + \frac{1}{2} = 0

What is sqrt(4-x^2)>x+7 ?
The solution to sqrt(4-x^2)>x+7 is No Solution for x\in\mathbb{R}