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

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

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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
[Reveal] Spoiler: OA

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Last edited by Bunuel on 25 Jul 2017, 19:39, edited 1 time in total.
Renamed the topic and edited the question.

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Re: The number can be expressed as (63 - 36*3^(1/2))^(1/2) can be expresse [#permalink]

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

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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Re: The number can be expressed as (63 - 36*3^(1/2))^(1/2) can be expresse [#permalink]

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New post 26 Jul 2017, 08:10
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carcass wrote:
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



\(\sqrt{63-36\sqrt{3}}\) 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.


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Saquib
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Re: The number can be expressed as (63 - 36*3^(1/2))^(1/2) can be expresse [#permalink]

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New post 26 Jul 2017, 09:05
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carcass wrote:
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


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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Re: The number can be expressed as (63 - 36*3^(1/2))^(1/2) can be expresse [#permalink]

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New post 09 Aug 2017, 12:34
carcass wrote:
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


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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Re: The number can be expressed as (63 - 36*3^(1/2))^(1/2) can be expresse [#permalink]

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New post 19 Nov 2017, 07:35
EgmatQuantExpert wrote:
carcass wrote:
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



\(\sqrt{63-36\sqrt{3}}\) 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

Register for our Free Session on Number Properties (held every 3rd week) to solve exciting 700+ Level Questions in a classroom environment under the real-time guidance of our Experts :)

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Elegant solution!
But a bit hard to deduct under time pressure

Kudos [?]: 2 [0], given: 42

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Location: Kazakhstan
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GPA: 3.2
Re: The number can be expressed as (63 - 36*3^(1/2))^(1/2) can be expresse [#permalink]

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New post 19 Nov 2017, 07:48
VeritasPrepKarishma wrote:
carcass wrote:
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


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)

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?

Kudos [?]: 2 [0], given: 42

Re: The number can be expressed as (63 - 36*3^(1/2))^(1/2) can be expresse   [#permalink] 19 Nov 2017, 07:48
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