Verified answer

So we want to square the entire quantity

(√(3+2√2)-√(3-2√2))^2

(3+2√2)-√(9-8)-√(9-8)+(3-2√)

3+2√2-1-1+3-2√2

4

An alternative would be to do severe rationalization and simplification on the original and discover it equals 2, but I wouldn't have thought it came out so decently.

√(3 + 2√2) =√2+1 and √(3 - 2√2) =√2-1.

So x = 2 and x^2 = 4. SImple!

x = √(3 + 2√2) - √(3 - 2√2)

x^2= {3 + 2(2^1/2)} - {3 - 2(2^1/2)}

= 3 + 2(2^1/2) - 3 + 2(2^1/2)

= 2(2^1/2) + 2(2^1/2)

= (4*2) ^1/2 + (4*2) ^1/2

= 8^1/2 + 8^1/2

= 16^1/2

= (4*4)^1/2

ans.

x^2 = 4

x^2 = (sqrt(3 + 2*sqrt(2)) - sqrt(3 - 2*sqrt(2)))^2

x^2 = 4

x1 = + 2

x2 = - 2