# Need the derivation/proof of the method

1. Jun 4, 2016

### Chaos_Enlightened

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. Jun 5, 2016

### 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"?

3. Jun 5, 2016

### Chaos_Enlightened

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. Jun 5, 2016

### Mastermind01

A quadratic surd can be written as $a + \sqrt(b)$ 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
$\sqrt(x) + \sqrt(y)$. Thus you equate $a + \sqrt(b)$ $(\sqrt(x) + \sqrt(y))^2$ . Compare terms to get the value of a and b.

You can also think of it this way. We know that $(m + n)^2 = m^2 + n^2 + 2mn$ . Since $a + \sqrt(b)$ is a square it must be of the form $m^2 + n^2 + 2mn$ . 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
5. Jun 5, 2016

### Chaos_Enlightened

Thank you mastermind01 it was helpful