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Bunuel
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Ans : B

B can be written as = (( 7x - 6y) ^2)^1/2 = 7x - 6y = integer
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its a direct application of formula...
(a+b)^2 = a^2 + 2ab + b^2

(a-b)^2 = a^2 -2ab + b^2

ans B
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Is there an approach to this if one does not immediately recognize (x^2 - 2xy +y^2) ?
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Bunuel
For integers x and y, which of the following MUST be an integer?


A. \(\sqrt{25x^2+30xy+36y^2}\)

B. \(\sqrt{49x^2−84xy+36y^2}\)

C. \(\sqrt{16x^2−y^2}\)

D. \(\sqrt{64x^2−64xy−64y^2}\)

E. \(\sqrt{81x^2+25xy+16y^2}\)

If the term inside the square root is a perfect square, then the whole term will be an integer.
We need to try and write the terms in the for of x^2 +- 2ax + a^2

A. \(\sqrt{25x^2+30xy+36y^2}\) = \(\sqrt{25x^2+5*6 xy+36y^2}\) this cannot be a perfect square

B. \(\sqrt{49x^2−84xy+36y^2}\) = \(\sqrt{49x^2−2*6*7*xy+36y^2}\) = \(\sqrt{(7x−6y)^2}\)
This is a perfect square.

Correct Option: B
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The options are the expanded form of identities which are commonly tested on GMAT.

B is the answer and the identity used in B is: (a-b)^2= a^2+b^2-2ab
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Bunuel
For integers x and y, which of the following MUST be an integer?


A. \(\sqrt{25x^2+30xy+36y^2}\)

B. \(\sqrt{49x^2−84xy+36y^2}\)

C. \(\sqrt{16x^2−y^2}\)

D. \(\sqrt{64x^2−64xy−64y^2}\)

E. \(\sqrt{81x^2+25xy+16y^2}\)

We need to determine which of the answer choices must be an integer.

Let’s take a look at answer choice B:

√(49x^2 - 84xy + 36y^2)

√(7x - 6y)(7x - 6y)

√(7x - 6y)^2 = |7x - 6y|

Since x and y are integers, |7x - 6y| is an integer.

Answer: B
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Can we not apply the same rule to options A and D?
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I mean option A and E
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Bunuel
For integers x and y, which of the following MUST be an integer?

A. \(\sqrt{25x^2+30xy+36y^2}\)

B. \(\sqrt{49x^2−84xy+36y^2}\)

C. \(\sqrt{16x^2−y^2}\)

D. \(\sqrt{64x^2−64xy−64y^2}\)

E. \(\sqrt{81x^2+25xy+16y^2}\)
santro789
Can we not apply the same rule to options A and D E?
santro789 , I'm not sure which rule you mean. In fact, I'm slightly confused by your question. Under either rule, how do you see A and E as possible answers?

OptimusPrepJanielle wrote
Quote:
If the term inside the square root is a perfect square, then the whole term will be an integer.
We need to try and write the terms in the form of x^2 +- 2ax + a^2
ScottTargetTestPrep wrote
Quote:
√(7x - 6y)^2 = |7x - 6y|

Since x and y are integers, |7x - 6y| is an integer.
The answer to your question is no. The most basic reason: the middle term in both A and E prevents both from being perfect squares.

If A were a perfect square, looking at its terms' coefficients, it would be
\(\sqrt{(5x + 6y)^2}\)= \(\sqrt{25x^2 + 60xy + 36y^2}\)

Answer A's middle term is 30xy, not 60xy. That means it's not a perfect square.

Answer E has the same problem. Looking at its coefficients, if it were a perfect square it would be
\(\sqrt{(9x + 4y)^2}\\
=\sqrt{81x^2 + 72xy + 16y^2}\)

The middle term in E is 25xy, not 72xy. In neither A nor E can we get a perfect square under the square root sign. If we could, we would get an integer: the square root of a perfect square is an integer. Just think about a couple of numeric values: \(\sqrt{2^2} = 2\), and \(\sqrt{41^2} = 41\)

Without being able to factor A and E into \((a + b)^2\) or \((a - b)^2\) because their middle terms prevent them from being perfect squares, we certainly cannot use absolute value analysis to prove what they are not.

Hope that helps.
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