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29 Jun 2024, 11:51
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Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
75%
(hard)
Question Stats:
60% (02:31) correct
40% (00:49) wrong based on 5 sessions
History
Date
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The integer v is greater than 1. If v is the square of an integer, which of the following numbers must also be the square of an integer?
Indicate all such numbers.
A. 81v
B. \(25v+10\sqrt{v}+1\)
C. \(4v^2+4\sqrt{v}+1\)
Indicate all such numbers.
A. 81v
B. \(25v+10\sqrt{v}+1\)
C. \(4v^2+4\sqrt{v}+1\)
13 Nov 2024, 06:01
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OE
If v is the square of an integer, then \(\sqrt{v}\) is an integer. You can use this fact, together with the fact that the product and the sum of integers are also integers, to examine the first two choices.
Choice A: The positive square root of 81v is \(9\sqrt{v}\), which is an integer. So 81v is the square of an integer.
Choice B: \( 25v + 10\sqrt{v} + 1 = \) \( (5 \sqrt{v}+1\) and \( + 1)^2 \) and \(5\sqrt{v}+1\) is an integer. So \(25v + 10\sqrt{v} + 1\) is the square of an integer.
Choice C: Since there is no obvious way to factor the given expression, you may suspect that it is not the square of an integer. To show that a given statement is not true, it is sufficient to find one counterexample. In this case, you need to find one value of v such that v is the square of an integer but \(4v^2+4\sqrt{v}+1\) is not the square of an integer. If v = 4, then \(4v^2+4\sqrt{v}+1\) = 64 + 8 + 1 = 73, which is not the square of an integer. This proves that \(4v^2+4\sqrt{v}+1\) does not have to be the square of an integer.
The correct answer consists of Choices A and B.
If v is the square of an integer, then \(\sqrt{v}\) is an integer. You can use this fact, together with the fact that the product and the sum of integers are also integers, to examine the first two choices.
Choice A: The positive square root of 81v is \(9\sqrt{v}\), which is an integer. So 81v is the square of an integer.
Choice B: \( 25v + 10\sqrt{v} + 1 = \) \( (5 \sqrt{v}+1\) and \( + 1)^2 \) and \(5\sqrt{v}+1\) is an integer. So \(25v + 10\sqrt{v} + 1\) is the square of an integer.
Choice C: Since there is no obvious way to factor the given expression, you may suspect that it is not the square of an integer. To show that a given statement is not true, it is sufficient to find one counterexample. In this case, you need to find one value of v such that v is the square of an integer but \(4v^2+4\sqrt{v}+1\) is not the square of an integer. If v = 4, then \(4v^2+4\sqrt{v}+1\) = 64 + 8 + 1 = 73, which is not the square of an integer. This proves that \(4v^2+4\sqrt{v}+1\) does not have to be the square of an integer.
The correct answer consists of Choices A and B.
