i*sqrt(6) for complex numbers (no solution for real numbers)
2007-01-04 07:46:21
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answer #1
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answered by supersonic332003 7
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Hi nick-
To take the square root of a negative number, mathematicians designated 'i' to be â-1. Numbers including an 'i' are called "imaginary". Using this, we can use the rule that âa ·âb = âab to find â-6.
We can say that â-6 is equal to â-1 · â6, and so â-6= iâ6. A general rule for finding â-a is that it is equal to iâa.
Hope this helped!
2007-01-04 15:47:22
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answer #2
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answered by Texas Cowgirl 3
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the answer is i times square root of 6. it's a complex number because i stands for the square root of negative numbers.
2007-01-04 15:50:09
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answer #3
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answered by sWtnsiMpLe 3
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it's an imaginary number. You can't have a rational square root of a negative, since if you multiply two negs you get a positive, and two positives also yields a positive, and to be a root, they need to be equal. If you had an imaginary "half negative" or something like that, you might be able to have one.
2007-01-04 15:46:31
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answer #4
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answered by loon_mallet_wielder 5
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By definition, i^2 = -1 => i = +/- sqrt -1 (keep positive value only)
=> i = sqrt -1
then sqrt -6 = sqrt (-1 x 6) = sqrt -1 x sqrt 6 = i x sqrt 6 =i x 2.45
Answer : 2.45i
An other example: sqrt -64 ?
sqrt -64 = sqrt (64 x -1) = sqrt 64 x sqrt -1 = 8 X i = 8i
In general : sqrt -a = sqrt(a x -1) =sqrt a x sqrt -1 = sqrt a x i
= i sqrt a
2007-01-04 16:33:50
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answer #5
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answered by frank 7
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in real number, it is not possible to square root a negative number. However, you can do in in complex number.
sqr(-16)= 4i and -4i
2007-01-04 15:47:18
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answer #6
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answered by 7
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The answer is an imaginary number
i(6)
the squareroot is in parentheses
2007-01-04 15:47:45
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answer #7
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answered by Anonymous
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