Need the derivation/proof of the method

  • #1

Homework Statement


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


Homework Equations




The Attempt at a Solution


I'm stumped how can we take √x+√y
 

Answers and Replies

  • #2
34,695
6,399

Homework Statement


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


Homework Equations




The Attempt at a Solution


I'm stumped how can we take √x+√y
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"?
 
  • #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?)
 
  • #4
201
51
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).
 
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  • #5
Thank you mastermind01 it was helpful
 
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