Question

Find the solution B=√2+√3-√2-√3

Answer 1

b= 0 because you take the square root of 2 + the square root of 3 and then subtract that by the square root of 2 and then subtract that by the square root of 3 it will give you zero

Answer 2

`B=√(2)`

Explanation:

Original equation:

`B = \sqrt{2+√(3)} - \sqrt{2 - √(3)}\\`

Square both sides:

`B^2=(\sqrt{2+√(3)})^2 + 2(\sqrt{2+√(3)} * (-\sqrt{2 - √(3)}) + (-\sqrt{2-√(3)})^2`

Cancel out the square roots and squares:

`B^2=2+√(3)}+ 2(\sqrt{2+√(3)} * (-\sqrt{2 - √(3)}) + 2-√(3)`

Add the sqrt(3) and -sqrt(3) as well as 2 and 2

`B^2=4 + 2(\sqrt{2+√(3)} * (-\sqrt{2 - √(3)})`

Use the identity:
`√(a) * √(b) = √(a * b)` to rewrite the two square roots being multiplied:

`B^2=4 + 2(-\sqrt{(2+√(3)) * (2 - √(3)}))`

Use difference of squares:
`(a-b)(a+b) = a^2-b^2`

`B^2=4 + 2(-\sqrt{2^2-√(3)^2})`

Square both sides

`B^2=4 + 2(-√(4-3))`

Subtract:

`B^2=4 + 2(-√(1))`

Evaluate square root of 1:

`B^2=4 + 2(-1)`

Multiply

`B^2=4 -2`

Subtract

`B^2=2`

Take square root of both sides:

`B=√(2)`