Bunuel wrote:

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 wrote:

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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