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If \(a\) and \(b\) are integers such that \(a < b < 0\) and \(a^2 = b^2 + 7\), what is the value of \(a-b\)?
A. -7
B. -4
C. -3
D. -1
E. 1
A. -7
B. -4
C. -3
D. -1
E. 1
09 May 2024, 00:35
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Official Solution:
If \(a\) and \(b\) are integers such that \(a < b < 0\) and \(a^2 = b^2 + 7\), what is the value of \(a-b\)?
A. -7
B. -4
C. -3
D. -1
E. 1
Given: \(a^2=b^2+7\).
Rearrange and apply \(a^2 - b^2 = (a - b)(a + b)\) to get \((a - b)(a + b) = 7\). Since \(a\) and \(b\) are integers, both \(a + b\) and \(a - b\) are integers as well. Thus, we have the product of two integers equal to 7. There are only two combinations of such factors possible: (1, 7) and (-1, -7). Given that both \(a\) and \(b\) are negative, the first case is not possible, so \(a - b\) must be either -1 or -7. However, it cannot be -7 because in this case, \(a + b\) must be -1, and no two negative integers yield a sum of -1. Therefore, \(a - b = -1\).
Answer: D
If \(a\) and \(b\) are integers such that \(a < b < 0\) and \(a^2 = b^2 + 7\), what is the value of \(a-b\)?
A. -7
B. -4
C. -3
D. -1
E. 1
Given: \(a^2=b^2+7\).
Rearrange and apply \(a^2 - b^2 = (a - b)(a + b)\) to get \((a - b)(a + b) = 7\). Since \(a\) and \(b\) are integers, both \(a + b\) and \(a - b\) are integers as well. Thus, we have the product of two integers equal to 7. There are only two combinations of such factors possible: (1, 7) and (-1, -7). Given that both \(a\) and \(b\) are negative, the first case is not possible, so \(a - b\) must be either -1 or -7. However, it cannot be -7 because in this case, \(a + b\) must be -1, and no two negative integers yield a sum of -1. Therefore, \(a - b = -1\).
Answer: D
25 May 2024, 01:17
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a^2 - b^2 = 7
Given both a and b are negative and a < b
(-4)^2 - (-3)^2 = 7
16 - 9 = 7
a - b = - 4 - (-3) = -1
Given both a and b are negative and a < b
(-4)^2 - (-3)^2 = 7
16 - 9 = 7
a - b = - 4 - (-3) = -1
