Need the derivation/proof of the method

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The discussion focuses on the method for finding the square root of a quadratic surd, specifically the expression a + √b. It explains that this can be equated to the form √x + √y, leading to the equations a = x + y and b = 2√xy after squaring both sides. Participants clarify terminology, confirming that "quadric surds" should be "quadratic surds." The rationale behind using the form √x + √y is that it allows for the separation of rational and irrational components in the equation. Overall, the method relies on comparing terms after squaring to derive the values of a and b.
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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
 
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Chaos_Enlightened said:

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"?
 
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?)
 
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).
 
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Thank you mastermind01 it was helpful
 
I tried to combine those 2 formulas but it didn't work. I tried using another case where there are 2 red balls and 2 blue balls only so when combining the formula I got ##\frac{(4-1)!}{2!2!}=\frac{3}{2}## which does not make sense. Is there any formula to calculate cyclic permutation of identical objects or I have to do it by listing all the possibilities? Thanks
Since ##px^9+q## is the factor, then ##x^9=\frac{-q}{p}## will be one of the roots. Let ##f(x)=27x^{18}+bx^9+70##, then: $$27\left(\frac{-q}{p}\right)^2+b\left(\frac{-q}{p}\right)+70=0$$ $$b=27 \frac{q}{p}+70 \frac{p}{q}$$ $$b=\frac{27q^2+70p^2}{pq}$$ From this expression, it looks like there is no greatest value of ##b## because increasing the value of ##p## and ##q## will also increase the value of ##b##. How to find the greatest value of ##b##? Thanks
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