sqrt question
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04-06-2017, 04:01 PM
(This post was last modified: 04-06-2017 04:19 PM by pier4r.)
Post: #2
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RE: sqrt question
literally asking google (I guess they do not have enough input checks).
![]() After some more thinking, I guess google is quite right. I was fooled at first. \( \sqrt{-4} \cdot \sqrt{-9} = \sqrt{4 i^2 } \cdot \sqrt{9 i^2 } = \sqrt{36 i^4 } = 6i^2 = -6 \) refreshing the operators precedence: https://en.wikipedia.org/wiki/Order_of_operations . The root come firsts, so "6" is not possible (that is, taking away the minus sign because -1*-1 = 1 , if one does not consider it as iota squared). Wikis are great, Contribute :) |
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Messages In This Thread |
sqrt question - KeithB - 04-06-2017, 03:25 PM
RE: sqrt question - pier4r - 04-06-2017 04:01 PM
RE: sqrt question - Namir - 04-06-2017, 04:02 PM
RE: sqrt question - KeithB - 04-06-2017, 04:51 PM
RE: sqrt question - Han - 04-06-2017, 05:46 PM
RE: sqrt question - pier4r - 04-06-2017, 05:15 PM
RE: sqrt question - KeithB - 04-06-2017, 06:03 PM
RE: sqrt question - Han - 04-06-2017, 06:18 PM
RE: sqrt question - Claudio L. - 04-07-2017, 01:23 PM
RE: sqrt question - Han - 04-07-2017, 04:48 PM
RE: sqrt question - Claudio L. - 04-07-2017, 09:15 PM
RE: sqrt question - Han - 04-07-2017, 10:54 PM
RE: sqrt question - Claudio L. - 04-09-2017, 03:59 AM
RE: sqrt question - David Hayden - 04-24-2017, 09:36 PM
RE: sqrt question - Claudio L. - 04-26-2017, 03:08 AM
RE: sqrt question - Han - 04-28-2017, 06:09 PM
RE: sqrt question - nsg - 04-07-2017, 11:34 PM
RE: sqrt question - Vtile - 04-09-2017, 10:41 AM
RE: sqrt question - nsg - 04-09-2017, 05:26 PM
RE: sqrt question - Vtile - 04-09-2017, 11:07 PM
RE: sqrt question - nsg - 04-10-2017, 01:44 AM
RE: sqrt question - Vtile - 04-25-2017, 11:38 PM
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