What is sqrt(b+5)+sqrt(b+10)>2 ?

The solution to sqrt(b+5)+sqrt(b+10)>2 is b>=-5

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sqrt(b+5)+sqrt(b+10)>2

b + 5 + b + 10 > 2

b \geq - 5

Interval Notation

[- 5, \infty)

Solution steps

b + 5 + b + 10 > 2

Subtract

b + 10

from both sides

b + 5 + b + 10 - b + 10 > 2 - b + 10

Simplify

b + 5 > 2 - b + 10

f (x) > g (x)

then

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

For

2 - b + 10 < 0 : b > - 6

For

2 - b + 10 \geq 0 :

No Solution

Find singularity points

Find non-negative values for radicals:

b \geq - 5

Combine the intervals

(No Solution or b > - 6) and b \geq - 5

Merge Overlapping Intervals

b \geq - 5

Verify Solutions:

b \geq - 5

True

The solution is

b \geq - 5

4 (- 5 x + 1) = 64

z^{\frac{3}{4}} \cdot z^{2}

- 11 + 10 (p + 10) = 4 - 5 (2 p + 11)

(\frac{9}{2} x^{\frac{1}{2}})^{2}

x^{2} - x - 12 = 0

What is sqrt(b+5)+sqrt(b+10)>2 ?
The solution to sqrt(b+5)+sqrt(b+10)>2 is b>=-5