Let y = √(1 + √(-3)) + √(1 - √(-3))
so y^2 = (√(1 + √(-3)) + √(1 - √(-3)))^2
= (1 + √(-3)) + (1 - √(-3)) + 2 √((1 + √(-3))(1 - √(-3)))
= (1 + √(-3)) + (1 - √(-3)) + 2 √(1 - (-3)))
= 6
so y = √(6) as LHS is positive
hence √(1 + √(-3)) + √(1 - √(-3)) = √(6)
2 comments:
How do you get from
This (√(1 + √(-3)) + √(1 - √(-3)))^2
to this? (1 + √(-3)) + (1 - √(-3)) + 2 √((1 + √(-3))(1 - √(-3)))
How do you come to that step? Can you show me each little part squaring the top term?
using (a+b)^2= a ^2 + b^2 + 2ab
and a= √(1 + √(-3)
b = √(1 - √(-3)))
you get a^2 = 1 + √(-3)
b^2 = 1 - √(-3)
2ab = 2 √((1 + √(-3))(1 - √(-3)))
is it not obvious
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