Directions: For the following question, select all the answer choices that apply.Q. 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)81vb)25v +10 √v +1c)4v2 + 4 √v +1Correct answer is option 'A,B'. Can you explain this answer?

Question

Answer

If v is the square of an integer, then √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 √v, which is an integer. So 81v is the square of an integer.

Choice B: 25v + 10 √v + 1 = (5 √v + 1)² and 5√v + 1 is an integer. So 25v + 10^√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 4v² + 4√v + 1 is not the square of an integer. If v = 4, then 4v² + 4√v + 1 = 64 + 8 + 1 = 73, which is not the square of an integer. This proves that 4v² + 4√v + 1 does not have to be the square of an integer.

The correct answer consists of Choices A and B.

Answer

If v is the square of an integer, then √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 √v, which is an integer. So 81v is the square of an integer.

Choice B: 25v + 10 √v + 1 = (5 √v + 1)² and 5√v + 1 is an integer. So 25v + 10^√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 4v² + 4√v + 1 is not the square of an integer. If v = 4, then 4v² + 4√v + 1 = 64 + 8 + 1 = 73, which is not the square of an integer. This proves that 4v² + 4√v + 1 does not have to be the square of an integer.

The correct answer consists of Choices A and B.

Answer

Identification of Required Numbers:
- The question states that v is an integer greater than 1 and is the square of an integer.
- We need to determine which of the given numbers must also be the square of an integer based on this information.

Analysis of Answer Choices:
a) 81v
- Since v is the square of an integer, multiplying v by 81 will still result in a perfect square because 81 is the square of 9. Thus, 81v is the square of an integer.
b) 25v + 10√v + 1
- The term 25v is the square of an integer if v is the square of an integer. However, the other terms 10√v + 1 may not necessarily result in a perfect square. Therefore, this expression may or may not be the square of an integer.
c) 4v^2 + 4√v + 1
- The term 4v^2 is the square of an integer if v is the square of an integer. However, the other terms 4√v + 1 may not necessarily result in a perfect square. Therefore, this expression may or may not be the square of an integer.

Final Selection:
- Based on the analysis, the numbers that must be the square of an integer are:
- 81v
- 25v + 10√v + 1
Therefore, the correct answer choices are options A and B.

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