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.