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Need the derivation/proof of the method

  1. Jun 4, 2016 #1
    1. The problem statement, all variables and given/known data
    The method given in my book is to find the square root of a quadric surds is to :
    Consider a and b where both are rational and √b is a surd
    Equate the square root of a+√b to √x+√y
    ie (a+√b)^(1/2)=√x+√y
    Squaring both sides we get a=x+y and b=2√xy


    2. Relevant equations


    3. The attempt at a solution
    I'm stumped how can we take √x+√y
     
  2. jcsd
  3. Jun 5, 2016 #2

    Mark44

    Staff: Mentor

    I don't understand your question here (in part 3).

    Is there more to this method than you have shown? Does you book say anything more about x and y?

    Also, when you write "the square root of a quadric surds" do you mean "quadratic surd"?
     
  4. Jun 5, 2016 #3
    Yes I mean quadratic (mistake) and I mean that I don't understand where does the √x+√y come from (are we considering the value of (a+√b)^(1/2) as √x +√y?)
     
  5. Jun 5, 2016 #4
    A quadratic surd can be written as [itex] a + \sqrt(b) [/itex] and the problem is to find the square root of the surd. The quadratic surd will be the square of some number of the form
    [itex] \sqrt(x) + \sqrt(y) [/itex]. Thus you equate [itex] a + \sqrt(b) [/itex] [itex] (\sqrt(x) + \sqrt(y))^2 [/itex] . Compare terms to get the value of a and b.

    You can also think of it this way. We know that [itex] (m + n)^2 = m^2 + n^2 + 2mn [/itex] . Since [itex] a + \sqrt(b) [/itex] is a square it must be of the form [itex] m^2 + n^2 + 2mn [/itex] . We are dealing only with quadratic surds here so 2mn must be the irrational part of the square and m^2 + n^2 must be the rational part (since m and n can be at most quadratic surds).
     
    Last edited: Jun 5, 2016
  6. Jun 5, 2016 #5
    Thank you mastermind01 it was helpful
     
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