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17 Jul 2017, 04:29
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If \(x\) and \(y\) are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?
(1) \(|xy|\) is NOT a square of an integer
(2) Point \((x, y)\) is above x-axis
(1) \(|xy|\) is NOT a square of an integer
(2) Point \((x, y)\) is above x-axis
17 Jul 2017, 04:30
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Official Solution:
If \(x\) and \(y\) are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?
This question is quite challenging, and it requires careful attention to detail. It is recommended to read the solution thoroughly and pay close attention to every detail.
To begin with, the inequality \(-x \leq y \leq x\) has two implications:
1. It ensures that \(x^2-y^2\) is non-negative, which means that the square root of this number will be well-defined.
2. It guarantees that \(x+y\) is non-negative, which means that the square root cannot be a negative number.
Moving forward, since \(-x \leq x\), we can infer that \(x\) is non-negative (\(x \geq 0\)).
Next, before delving into the statements, let's rephrase the question as follows:
Does \(\sqrt{x^2 - y^2} = x + y\)?
Square both sides: does \(x^2 - y^2 = x^2+2xy + y^2\)?
Does \(xy+y^2=0\)? Notice here that we cannot reduce this by \(y\), because we'll loose a possible root: \(y=0\).
Does \(y(x+y)=0\)?
Does \(y=0\) or \(x=-y\)?
(1) \(|xy|\) is NOT a square of an integer.
If \(y = 0\) were true, then \(|xy|\) would be 0, which is a square of an integer.
If \(x = -y\) were true, then \(|xy|\) would be \(y^2\), which is a square of an integer (since we are told that \(y\) is an integer).
Therefore, since we are told that \(|xy|\) is NOT a square of an integer, we can conclude that neither \(y=0\) nor \(x=-y\) is true. Sufficient.
(2) Point \((x, y)\) is above x-axis
If \(y = 0\) were true, then point \((x, y)\) would be ON the x-axis.
If \(x = -y\) were true, then then point \((x, y)\) would be \((x, -x)\), so would be (non-negative, non-positive), which would result in the point lying either on the x-axis or below it.
Therefore, since we are told that the point \((x, y)\) is above x-axis, we can conclude that neither \(y=0\) nor \(x=-y\) is true. Sufficient.
Answer: D
If \(x\) and \(y\) are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?
This question is quite challenging, and it requires careful attention to detail. It is recommended to read the solution thoroughly and pay close attention to every detail.
To begin with, the inequality \(-x \leq y \leq x\) has two implications:
1. It ensures that \(x^2-y^2\) is non-negative, which means that the square root of this number will be well-defined.
2. It guarantees that \(x+y\) is non-negative, which means that the square root cannot be a negative number.
Moving forward, since \(-x \leq x\), we can infer that \(x\) is non-negative (\(x \geq 0\)).
Next, before delving into the statements, let's rephrase the question as follows:
Does \(\sqrt{x^2 - y^2} = x + y\)?
Square both sides: does \(x^2 - y^2 = x^2+2xy + y^2\)?
Does \(xy+y^2=0\)? Notice here that we cannot reduce this by \(y\), because we'll loose a possible root: \(y=0\).
Does \(y(x+y)=0\)?
Does \(y=0\) or \(x=-y\)?
(1) \(|xy|\) is NOT a square of an integer.
If \(y = 0\) were true, then \(|xy|\) would be 0, which is a square of an integer.
If \(x = -y\) were true, then \(|xy|\) would be \(y^2\), which is a square of an integer (since we are told that \(y\) is an integer).
Therefore, since we are told that \(|xy|\) is NOT a square of an integer, we can conclude that neither \(y=0\) nor \(x=-y\) is true. Sufficient.
(2) Point \((x, y)\) is above x-axis
If \(y = 0\) were true, then point \((x, y)\) would be ON the x-axis.
If \(x = -y\) were true, then then point \((x, y)\) would be \((x, -x)\), so would be (non-negative, non-positive), which would result in the point lying either on the x-axis or below it.
Therefore, since we are told that the point \((x, y)\) is above x-axis, we can conclude that neither \(y=0\) nor \(x=-y\) is true. Sufficient.
Answer: D
General Discussion
17 Jul 2017, 06:23
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Bunuel
Hi Bunuel,
Can we say that x = -y is the same as -x = y in this case? And if that is the case, will that affect the fact that statement (2) is sufficient?
Thanks
