What is x+2<sqrt(10-x^2) ?

The solution to x+2<sqrt(10-x^2) is -sqrt(10)<= x<1

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x+2<sqrt(10-x^2)

x + 2 < 10 - x^{2}

- 10 \leq x < 1

Interval Notation

[- 10, 1)

Decimal Notation

- 3.16227 \dots \leq x < 1

Solution steps

x + 2 < 10 - x^{2}

Switch sides

10 - x^{2} > x + 2

f (x) > g (x)

then

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

For

x + 2 < 0 : x < - 2

For

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

Find singularity points

Find non-negative values for radicals:

- 10 \leq x \leq 10

Combine the intervals

(- 2 \leq x < 1 or x < - 2) and - 10 \leq x \leq 10

Merge Overlapping Intervals

- 10 \leq x < 1

Verify Solutions:

- 10 \leq x < 1

True

The solution is

- 10 \leq x < 1

1 - 27 a^{3} b^{3}

a^{2} - 10 a + 21

4 (x - 5)^{2} - 50 = 2

8 e + 3 (5 - e) = 10

- \frac{x}{3} + 6 - 5 = 0