The number \(\sqrt{63-36\sqrt{3}}\) can be expressed as \(x + y \sqrt{3}\) for some integers x and y. What is the value of xy ?

A. –18

B. –6

C. 6

D. 18

E. 27
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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
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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?

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