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anonymous
 5 years ago
Could anyone please help me with simplifying radicals? 15
anonymous
 5 years ago
Could anyone please help me with simplifying radicals? 15

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0√(15) is the problem?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0what is your question?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it is odd and cannot be simplfied

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ok so if you break it into it's factors you get √(1)(3)(5) √(1) = i the rest cannot be simplified so you have i√15 for an answer

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0√49 is also odd but simplifies to 7 mmolina

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so the answer is 1/15

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0no, the answer is: i*srqr(15)

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i squared as in india type of i?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0hmm never heard of that before. what rule is that derived from?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0i is the imaginary unit √(1) because you cannot take the square root of a negative number.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0okay. let me try that...

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0set in you personal calculator \[\sqrt{1}\] this will give you an error as the answer. So every time you have an expresion like that you have to write an i in front of the root and make the number inside the root positive.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the entire problem is quadratic formulas which is 1 +/ \[\sqrt{15}/2\]

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0then there are no real solutions

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I have the left side down and the fraction beneathe, but complicated fixing right side. is that possible for a solution set? the answer must be two separate numbers

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0not always, quadratics have AT MOST 2 solutions but can have only one or no real solutions, this appears to be a classic no real solutions problem

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0the solutions to a quadratic are where the graph crossed the xaxis, if you graph the original problem does it cross the xaxis?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0there is no graphing, and there is no option for no solutionthere is indeed a solution. haha

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0It might make things simpler if you tell us what the original problem is.

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0it must form an equation with only addition and subraction signs between them

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I did if you look above

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0did you not understand it?

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I understand that you have simplified the original problem but if you post what the quadratic is then we could verify that your solutions are correct.
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