Erjan_S wrote:

Why the co-efficient of the irrational term has to be equal to on both sides? Is there such a rule?

You'd never need to mathematically prove that's true on the GMAT, but it is true. I can explain, but test takers don't need to understand how to do this in order to solve GMAT questions:

We arrived at this:

\(63 - 36\sqrt{3} = x^2 + 3y^2 + 2xy \sqrt{3}\)

x and y are integers, so x^2 + 3y^2 is just some integer, and 2xy is also some integer. So let's just let a = x^2 + 3y^2, and b = 2xy, to make this all easier to look at. Then a and b are integers and we have

\(63 - 36\sqrt{3} = a + b \sqrt{3}\)

Now the number √3 is what is called an 'irrational number', which means it is impossible to write √3 as some fraction c/d where c and d are both integers (I can prove that √3 is irrational separately if anyone is interested). Let's take the equation above, and write it with √3 on one side:

\(\begin{align}

63 - 36\sqrt{3} &= a + b \sqrt{3} \\

63 - a &= b \sqrt{3} + 36 \sqrt{3} \\

63 - a &= \sqrt{3} (b + 36) \\

\frac{63 - a}{b + 36} &= \sqrt{3}

\end{align}\)

But there's something wrong here: 63-a and b+36 are both integers. If this equation were right, then we would have just written √3 as a fraction using two integers, but we know that's impossible to do since √3 is an irrational number. So there must be something wrong with our solution, and the only thing that might be wrong is that we might have divided by zero in the second last line. So b+36 must be equal to 0 (and similarly 63-a must be equal to 0). From there we find that b = -36, so 2xy = -36, and xy = -18.

That's how you can prove that in similar equations, the rational parts and irrational parts must be equal. So if you saw, say, this equation:

5 + 10√2 = e + f + (m + p)√2

where all the letters are integers, then it would need to be true that e+f = 5, and m +p = 10.

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