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17 Jul 2017, 04:41
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E
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
95%
(hard)
Question Stats:
33% (03:36) correct
67%
(02:32)
wrong
based on 33
sessions
History
Date
Time
Result
Not Attempted Yet
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 NOT on x-axis.
(1) \(xy\) is NOT a square of an integer.
(2) Point (x, y) is NOT on x-axis.
17 Jul 2017, 04:41
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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 is indeed a complex question that requires a meticulous approach and careful reading of the solution.
First of all, \(-x \leq y \leq x\) guarantees two things:
1. \(x^2-y^2\geq 0\), ensuring that the square root of this number is defined.
2. \(x+y\geq 0\), so the square root will not equal a negative number.
Moreover, \(-x \leq x\) implies that \(x \geq 0\).
And finally, before moving to the statements, let's rephrase the question:
Does \(\sqrt{x^2 - y^2} = x + y\)?
Square both sides: does \(x^2 - y^2 = x^2+2xy + y^2\)?
Does \(xy+y^2=0\)? Note that we can't simplify this further by dividing by \(y\), as we will lose 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 \(y = 0\) was the case, then \(xy\) would be 0, which is a square of an integer. This contradicts the statement.
On the other hand, if \(x = -y\) and \(y ≠ 0\) were true, for instance if \(x = -y = 1\), then \(xy\) would be \(-x^2\), a negative value, which cannot be a square of an integer. So, \(x = -y\) IS a possibility.
However, there may be other solutions where \(x\) is not equal to \(-y\) and yet the statement could still hold true. For instance, \(x = 2\) and \(y =1\).
Thus, we have two different answers to the question: Not sufficient.
(2) Point \((x, y)\) is NOT on x-axis.
If \(y = 0\) was the case, then the point \((x, y)\) would be ON the x-axis. Therefore, we know that \(y ≠ 0\).
If \(x = -y\) and \(y ≠ 0\) were true, for instance if \(x = -y = 1\), then the point \((x, y)\) would be \((x, -x)\), so (positive, negative), which is a point below the x-axis. Thus, \(x = -y\) IS a possibility.
Similarly, \(x\) can be not equal to \(-y\), and the statement could still hold true. For example, \(x = 2\) and \(y =1\).
Thus, we have two different answers to the question: Not sufficient.
(1)+(2) We could still have the cases considered above: for instance, if \(x = -y\) and \(y ≠ 0\), such as when \(x = -y = 1\), the answer to our original question would be YES. However, if we consider a different case, like \(x = 2\) and \(y = 1\), the answer would be NO. Not sufficient.
Answer: E
If \(x\) and \(y\) are integers and \(-x \leq y \leq x\), does \(\sqrt{x^2 - y^2} = x + y\)?
This is indeed a complex question that requires a meticulous approach and careful reading of the solution.
First of all, \(-x \leq y \leq x\) guarantees two things:
1. \(x^2-y^2\geq 0\), ensuring that the square root of this number is defined.
2. \(x+y\geq 0\), so the square root will not equal a negative number.
Moreover, \(-x \leq x\) implies that \(x \geq 0\).
And finally, before moving to the statements, let's rephrase the question:
Does \(\sqrt{x^2 - y^2} = x + y\)?
Square both sides: does \(x^2 - y^2 = x^2+2xy + y^2\)?
Does \(xy+y^2=0\)? Note that we can't simplify this further by dividing by \(y\), as we will lose 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 \(y = 0\) was the case, then \(xy\) would be 0, which is a square of an integer. This contradicts the statement.
On the other hand, if \(x = -y\) and \(y ≠ 0\) were true, for instance if \(x = -y = 1\), then \(xy\) would be \(-x^2\), a negative value, which cannot be a square of an integer. So, \(x = -y\) IS a possibility.
However, there may be other solutions where \(x\) is not equal to \(-y\) and yet the statement could still hold true. For instance, \(x = 2\) and \(y =1\).
Thus, we have two different answers to the question: Not sufficient.
(2) Point \((x, y)\) is NOT on x-axis.
If \(y = 0\) was the case, then the point \((x, y)\) would be ON the x-axis. Therefore, we know that \(y ≠ 0\).
If \(x = -y\) and \(y ≠ 0\) were true, for instance if \(x = -y = 1\), then the point \((x, y)\) would be \((x, -x)\), so (positive, negative), which is a point below the x-axis. Thus, \(x = -y\) IS a possibility.
Similarly, \(x\) can be not equal to \(-y\), and the statement could still hold true. For example, \(x = 2\) and \(y =1\).
Thus, we have two different answers to the question: Not sufficient.
(1)+(2) We could still have the cases considered above: for instance, if \(x = -y\) and \(y ≠ 0\), such as when \(x = -y = 1\), the answer to our original question would be YES. However, if we consider a different case, like \(x = 2\) and \(y = 1\), the answer would be NO. Not sufficient.
Answer: E
11 Sep 2017, 07:25
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Can this question be solved by plugging in values?
the approach explained by bunuel is tough to comprehend does someone else has an easy method or trick for this question?
the approach explained by bunuel is tough to comprehend does someone else has an easy method or trick for this question?
