sqrt question

04062017, 04:01 PM
(This post was last modified: 04062017 04:19 PM by pier4r.)
Post: #2




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 :) 

« Next Oldest  Next Newest »

Messages In This Thread 
sqrt question  KeithB  04062017, 03:25 PM
RE: sqrt question  pier4r  04062017 04:01 PM
RE: sqrt question  Namir  04062017, 04:02 PM
RE: sqrt question  KeithB  04062017, 04:51 PM
RE: sqrt question  Han  04062017, 05:46 PM
RE: sqrt question  pier4r  04062017, 05:15 PM
RE: sqrt question  KeithB  04062017, 06:03 PM
RE: sqrt question  Han  04062017, 06:18 PM
RE: sqrt question  Claudio L.  04072017, 01:23 PM
RE: sqrt question  Han  04072017, 04:48 PM
RE: sqrt question  Claudio L.  04072017, 09:15 PM
RE: sqrt question  Han  04072017, 10:54 PM
RE: sqrt question  Claudio L.  04092017, 03:59 AM
RE: sqrt question  David Hayden  04242017, 09:36 PM
RE: sqrt question  Claudio L.  04262017, 03:08 AM
RE: sqrt question  Han  04282017, 06:09 PM
RE: sqrt question  nsg  04072017, 11:34 PM
RE: sqrt question  Vtile  04092017, 10:41 AM
RE: sqrt question  nsg  04092017, 05:26 PM
RE: sqrt question  Vtile  04092017, 11:07 PM
RE: sqrt question  nsg  04102017, 01:44 AM
RE: sqrt question  Vtile  04252017, 11:38 PM

User(s) browsing this thread: 1 Guest(s)