Saturday, April 21, 2012

show that √(1 + √(-3)) + √(1 - √(-3)) = √(6)

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)
at

2 comments:

  1. AnonymousApril 24, 2012 at 2:30 AM

    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?

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  2. kaliprasad April 24, 2012 at 8:22 PM

    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

    ReplyDelete