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If a and b are positive integers, a2 + b2 = 41, and a2 - b2 = 9, then
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07 Feb 2022, 07:00
07 Feb 2022, 07:00
Expert Reply
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Question Stats:
77% (01:30) correct
22% (01:06) wrong based on 27 sessions
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If a and b are positive integers, \(a^2 + b^2 = 41\), and \(a^2 - b^2 = 9\), then \(b = \)
(A) 3
(B) 4
(C) 5
(D) 6
(E) 7
(A) 3
(B) 4
(C) 5
(D) 6
(E) 7
Re: If a and b are positive integers, a2 + b2 = 41, and a2 - b2 = 9, then
[#permalink]
07 Feb 2022, 08:16
07 Feb 2022, 08:16
2
\(a^2 + b^2 = 41\) & \(a^2 - b^2 = 9\)
(a-b)(a+b) = 9
\(a\) & \(b\) both are \(> 0\)
Only one pair can be the solution.
\(a = 5\) & \(b = 4\)
Answer \(B\)
(a-b)(a+b) = 9
\(a\) & \(b\) both are \(> 0\)
Only one pair can be the solution.
\(a = 5\) & \(b = 4\)
Answer \(B\)
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Re: If a and b are positive integers, a2 + b2 = 41, and a2 - b2 = 9, then
[#permalink]
07 Feb 2022, 09:18
07 Feb 2022, 09:18
2
Carcass wrote:
If a and b are positive integers, \(a^2 + b^2 = 41\), and \(a^2 - b^2 = 9\), then \(b = \)
(A) 3
(B) 4
(C) 5
(D) 6
(E) 7
(A) 3
(B) 4
(C) 5
(D) 6
(E) 7
Take:
\(a^2 + b^2 = 41\)
\(a^2 - b^2 = 9\)
Subtract the bottom equation from the top equation to get: \((a^2 + b^2) - (a^2 - b^2) = 41 - 9\)
Simplify: \(2b^2 = 32\)
Divide both sides by 2 to get: \(b^2 = 16\)
So, EITHER \(b = 4\) OR \(b = -4\)
Since the question tells us that \(b\) is a positive integer, we can be certain that \(b = 4\)
Answer: B
