What is sqrt(b+12)-sqrt(b)>2 ?

The solution to sqrt(b+12)-sqrt(b)>2 is 0<= b<4

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

sqrt(b+12)-sqrt(b)>2

b + 12 - b > 2

0 \leq b < 4

Interval Notation

[0, 4)

Solution steps

b + 12 - b > 2

Add

b

to both sides

b + 12 - b + b > 2 + b

Simplify

b + 12 > 2 + b

f (x) > g (x)

then

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

For

2 + b < 0 :

No Solution

For

2 + b \geq 0 : b < 4

Find singularity points

Find non-negative values for radicals:

b \geq 0

Combine the intervals

(b < 4 or No Solution) and b \geq 0

Merge Overlapping Intervals

0 \leq b < 4

Verify Solutions:

0 \leq b < 4

True

The solution is

0 \leq b < 4

\frac{0.4 A ^{0.6}}{0.6 B ^{0.4}}

4 (x - 2) = 24

b - 7 = - 1

3 n^{2} - 12