SOLUTION: sqrt(3+sqrt(2)) + sqrt(3-sqrt(2)) = answer is sqrt(6+sqrt(28))

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Question 953761: sqrt(3+sqrt(2)) + sqrt(3-sqrt(2)) =
answer is sqrt(6+sqrt(28))

Answer by Theo(13342) About Me  (Show Source):
You can put this solution on YOUR website!
if you let a = sqrt(3+sqrt(2)) and if you let b = sqrt(3-sqrt(2)), and you let y = sqrt(6+sqrt(28)), then your equation becomes:

a + b = y

if you square both sides of the equation, you will get:

(a + b)^2 = y

if you simplify the equation, you will get:

a^2 + 2ab + b^2 = y^2

if you replace a with sqrt(3+sqrt(2)), you will get:

a^2 = (sqrt(3 + sqrt(2))^2 which becomes:

a^2 = 3 + sqrt(2))

similarly, if you replace b with sqrt(3-sqrt(2)), you will get:

b^2 = sqrt(3-sqrt(2))^2 which becomes:

b^2 = 3 - sqrt(2)).

so far you have:

a^2 = 3 + sqrt(2))
b^2 = 3 - sqrt(2))

your equation of a^2 + 2ab + b^2 = y^2 becomes:

3 + sqrt(2)) + 2ab + 3 - sqrt(2)) = y^2

now it's time to work on 2ab.

a is equal to sqrt(3+sqrt(2)) and b is equal to sqrt(3-sqrt(2)), so:

2ab is equal to 2 * a * b is equal to 2 * sqrt(3+sqrt(2)) * sqrt(3-sqrt(2)) which becomes:

2 * sqrt((3 + sqrt(2)) * (3 - sqrt(2)))

it may be hard to see, but what this is telling you you is that sqrt(k)*sqrt(m) is equal to sqrt(k*m) which is one of the arithmetic properties of square roots.

when you multiply (3 + sqrt(2)) * (3 - sqrt(2)), you will get:

(3 + sqrt(2)) * (3 - sqrt(2)) = 9 - 2 which is equal to 7.

therefore 2 * a * b = 2 * sqrt((3 + sqrt(2)) * (3 - sqrt(2))) becomes 2 * a * b = 2 * sqrt(7).

your equation of y^2 = (a + b)^2 = a^2 + 2ab + b^2 becomes:

y^2 = 3 + sqrt(2) + 2 * sqrt(7) + 3 - sqrt(2)

combine like terms to get:

y^2 = 6 + 2 * sqrt(7)

2 * sqrt(7) is equal to sqrt(4 * 7) which is equal to sqrt(28).

your equation becomes:

y^2 = 6 + sqrt(28)

take the square root of both sides of this equation to get:

y = sqrt(6 + sqrt(28))

the more simplified answer is y = sqrt(6 + 2 * sqrt(7)), but you were asked to duplicate the answer shown and so y = sqrt(6 + sqrt(28)) would be the one that you want.