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If $b$ is the square root of an integer,
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25 Nov 2025, 15:42
25 Nov 2025, 15:42
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Question Stats:
54% (01:24) correct
45% (01:35) wrong based on 37 sessions
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If $b$ is the square root of an integer, then which of the following numbers must be an integer?
Indicate all such numbers.
(A) $\(9 b^4-3 b^2+1\)$
(B) $\(\frac{b^2}{4}+4\)$
(c) $\(b^2+2 b+1\)$
Indicate all such numbers.
(A) $\(9 b^4-3 b^2+1\)$
(B) $\(\frac{b^2}{4}+4\)$
(c) $\(b^2+2 b+1\)$
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Re: If $b$ is the square root of an integer,
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13 Dec 2025, 13:15
13 Dec 2025, 13:15
Expert Reply
Let $\(b=\sqrt{n}\)$ where $n$ is an integer.
(A) $\(9 b^4-3 b^2+1\)$
$\(b^2=n, b^4=n^2\)$.
So $\(9 b^4-3 b^2+1=9 n^2-3 n+1\)$, which is an integer for any integer $n$.
So (A) must be an integer.
(B) $\(\frac{b^2}{4}+4\)$
$\(\frac{b^2}{4}+4=\frac{n}{4}+4\)$.
This is an integer only when $n$ is a multiple of 4 , which need not be true (e.g., $n=1$ gives $\(1 / 4+4\)$ ).
So $(B)$ is not guaranteed to be an integer.
(C) $\(b^2+2 b+1\)$
This equals $\((b+1)^2\)$.
With $\(b=\sqrt{n}\)$ and $n$ not necessarily a perfect square, $b$ need not be integer, so $\((b+1)^2\)$ need not be integer (e.g., $\(b=\sqrt{2}\)$ ).
So (C) is not guaranteed to be an integer.
Answer: (A) only.
(A) $\(9 b^4-3 b^2+1\)$
$\(b^2=n, b^4=n^2\)$.
So $\(9 b^4-3 b^2+1=9 n^2-3 n+1\)$, which is an integer for any integer $n$.
So (A) must be an integer.
(B) $\(\frac{b^2}{4}+4\)$
$\(\frac{b^2}{4}+4=\frac{n}{4}+4\)$.
This is an integer only when $n$ is a multiple of 4 , which need not be true (e.g., $n=1$ gives $\(1 / 4+4\)$ ).
So $(B)$ is not guaranteed to be an integer.
(C) $\(b^2+2 b+1\)$
This equals $\((b+1)^2\)$.
With $\(b=\sqrt{n}\)$ and $n$ not necessarily a perfect square, $b$ need not be integer, so $\((b+1)^2\)$ need not be integer (e.g., $\(b=\sqrt{2}\)$ ).
So (C) is not guaranteed to be an integer.
Answer: (A) only.
gmatclubot
Re: If $b$ is the square root of an integer, [#permalink]
13 Dec 2025, 13:15
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