The number can be expressed as (63 - 36*3^(1/2))^(1/2) can be expresse

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

Square both sides:

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

On the right, the integer part is the x^2 + 3y^2, so that equals 63, and the 2xy√3 is equal to -36√3, so 2xy = -36, and xy = -18.
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Here is what comes to my mind when I see this question:

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

Now, this is a PS question so I will have a unique value for xy. All I need to do is find one set of values which satisfy this equation.
There is nothing I can compare while there is the sqrt on the left hand side. So let's square both sides.

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

The co-efficient of the irrational term has to be equal to on both sides. So
-36 = 2xy
xy = -18

Answer (A)
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\(\sqrt{63-36\sqrt{3}}\) can be written as

=\(\sqrt{63-2*6*3\sqrt{3}}\)

=\(\sqrt{6^2 + 3\sqrt{3}^2 - 2*6*3\sqrt{3}}\)

=\(\sqrt{(6 - 3\sqrt{3})^2}\)

=\(6 - 3\sqrt{3}\)

Comparing it with \(x + y \sqrt{3}\)

We can say the value of x = 6 and y = -3

Thus, the product of x*y = -18

And the correct answer is Option A.


Thanks,
Saquib
Quant Expert
e-GMAT

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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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We can equate the two expressions:

√(63 - 36√3) = x + y√3

Squaring both sides we have:

63 - 36√3 = (x + y√3)^2

63 - 36√3 = (x + y√3)(x + y√3)

63 - 36√3 = x^2 + 2xy√3 + 3y^2

We can equate the terms containing √3 on each side of the equation. We see that:

-36√3 = 2xy√3

-18 = xy

Answer: A
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Why the co-efficient of the irrational term has to be equal to on both sides? Is there such a rule? Not in basic GMAT quant rules like GMAT official quantitative review...
I do have difficulties with this concept you have two sides of the equations with two terms on one side and 3 terms on the other side and then it looks like you arbitrarily pick a couple, how about other terms 63 and X^2 and 3y^2? Do they have to be equal and why?
VeritasPrepKarishma
Bunuel

Ian has already explained very well why this will be true.
I will just add why it makes intuitive sense to me.

An irrational number is one which cannot be constructed from ratio of integers. So I cannot have an expression with just integers equal to an irrational number.
\(\sqrt{3}\) cannot be (ab + c)/d etc etc where all are integers. There are no building blocks for irrational numbers among integers.

So if I have a \(\sqrt{3}\) on the left hand side, I must have one on the right hand side too.
If I have 4 \(\sqrt{3}\)s on the left hand side, I must have 4 \(\sqrt{3}\)s on the right hand side too. That is why I can equate the co-efficients.
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I think it's important to note that once we get to here:

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

The reason we can set \(-36 \sqrt{3}\) equal to \(2xy \sqrt{3}\) is because we know from the question that \(x\) and \(y\) are integers. If we didn't know that \(x\) and \(y\) were integers, then we would have more algebra to do and several more possibilities for potential solutions. For example, \(x = 3\) and \(y = \sqrt{3} - 3\) would be correct solutions to the equation.

Also, I think EgmatQuantExpert posted a very clever solution. However, I think parentheses are helpful around \(3\sqrt{3}\) in line 3. I have added those below:

1. \(\sqrt{63-36\sqrt{3}}\) can be written as

2. \(\sqrt{63-2*6*3\sqrt{3}}\)

3. \(\sqrt{6^2 + (3\sqrt{3})^2 - 2*6*3\sqrt{3}}\)

4. \(\sqrt{(6 - 3\sqrt{3})^2}\)

5. \(6 - 3\sqrt{3}\)

Hi IanStewart

Could you please help?

(63 - a) / (b + 36) = 3^1/2
integer 1 / integer 2 = 3^1/2
L.H.S cannot equal R.H.S since 3^1/2 is an irrational number. I get this part.

I don't understand what you mean by this, ?
So because the above case is impossible, the error we MIGHT have made is 0 / 0 = 3^1/2
undefined = 3^1/2 Is this what you are saying?

I mean I can certainly memorize this part, , but it wouldn't help improve my flair in solving tough GMAT Quant questions without knowing the logic behind it.

This is definitely true - if you want to answer harder GMAT questions, memorizing things is not all that helpful. You genuinely need to understand the mathematical concepts tested. Here, let's rewind a bit and look at some simpler algebra first. Say you solve an equation or equations, and end up with something nonsensical. That can mean one of two things:

• your equation was impossible to begin with. If you have x^2 + 1 = 0, for example, you can rewrite that x^2 = -1, which is impossible -- the problem is that the original equation cannot be true. Similarly if you try to solve simultaneously the two equations for parallel lines (e.g. y = 2x + 1 and y = 2x + 3) you end up with something nonsensical no matter what you do, because those two equations share no solutions; they can't both be true

• you did something illegal when solving the equation. When it comes to equations, the only common illegal thing you might do is accidentally divide by zero on both sides (when it comes to inequalities, on the other hand, there are a lot of illegal things one might be tempted to do). You can seemingly 'prove' anything is true if you allow yourself to divide by zero on both sides of an equation. For example, if we start with the equation x = 0, add x to both sides to get 2x = x, then divide by x on both sides, we get 2 = 1, which is clearly nonsense. There was nothing wrong with our original equation "x = 0"; instead the problem was that we divided by zero on both sides, when we divided by x, because x = 0.

That's what happened in my solution you quoted: we wrote √3 as a fraction with an integer numerator and integer denominator. That's nonsense, because √3 is an irrational number. The only way you get nonsense like that when solving an equation (an equation that wasn't nonsense to begin with) is if you divide by zero somewhere.
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\(\sqrt{\ 63-36\sqrt{\ 3}}=\ x+y\sqrt{3\ }\)

Hence when the number under the square root is squared we get :
\(63-36\sqrt{\ 3}=\ x^2+3y^2+2xy\sqrt{\ 3}\)
Since both x and y are integers we have :
\(x^2+3y^2\) to be an integer.
Hence the irrational term in the equation must be equated to the \(2xy\sqrt{\ 3}\)
\(2xy\sqrt{\ 3}=\ -36\sqrt{\ 3}\)
Hence xy = -18.

KarishmaB
A few questions:
-Does "can be expressed as" always mean equal to?
-Can you work backwards, using the answer choices to solve this question, and if so how would you do that?

Thank you

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