c) The square root of 9 and the imaginary square root of negative 1, which is i.
2007-11-26 07:34:51
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answer #1
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answered by Brian K² 6
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"the negative square root of negative nine"
- sqrt (-9)
first consider the sqrt ( -9)
sqrt (-9) = sqrt (-1) * sqrt(9)
since sqrt(ab) = sqrt(a) * sqrt(b)
by definition, i = sqrt(-1)
so sqrt (-9) = i * sqrt (9)
BUT ... sqrt (9) = 3 or -3
since 3*3 = 9 AND (-3)*(-3) = 9
so sqrt (-9) = 3i AND -3i
back to the original problem,
- sqrt (-9) = -3i AND 3i
both answers (c) and (d) are correct! (if the person who came up with the question and choice of answers did not realize this, then my guess is they are looking for the positive square root, 3i ... answer (c))
from a mathematical point of view, BOTH (c) and (d) are CORRECT.
added later: if you read the question differently ... it is asking for the negative square root of sqrt (-9) ... sqrt (-9) has the positive square root 3i and negative square root -3i
so in this case the answer is (d).
my final answer is (d) for this reason ... :-)
2007-11-26 15:49:21
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answer #2
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answered by Anonymous
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D
Square root of -9 is 3i but the negative square root means you have to attach a negative sign.
2007-11-26 15:39:23
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answer #3
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answered by The Oracle 2
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-9, because it is negative, cannot have a true square root. The value 'i' is used to represent the imaginary number which is defined as the square root of -1.
Therefore the answer would be C.
2007-11-26 15:40:34
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answer #4
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answered by Anonymous
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3i (it's 3 x square root of negative 1, which is i)
2007-11-26 15:40:18
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answer #5
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answered by strawberryfields 2
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a. 3
because you should take the square of 9 and then
the double negitives cancel eachother out so the answer is 3
2007-11-26 16:01:22
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answer #6
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answered by erinn m 1
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I´m sorry but between complex numbers there is only defined equal and not equal.
There do not exist positive or negative complex numbers as this implies > o<.Tell your teacher that this question is nonsense.
2007-11-26 15:47:01
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answer #7
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answered by santmann2002 7
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