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anonymous
 5 years ago
r^4 + 2r^2 + r = 0 ... How to solve this ???
anonymous
 5 years ago
r^4 + 2r^2 + r = 0 ... How to solve this ???

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anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0for the second term use factor theorm

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0note, if you put r=1 in the second bracket then it is zeroed out so (r+1) is a factor

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0therefore (r^3 +2r +1 ) = (r+1)(r^2 +ar ) let r=1 4 = 2(1+a) 1+a = 2 a= 1

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0umm if i put r = 1 in the second bracket the answer doesn't come out to be zero :S

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0ohh yeh, made mistake there

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0umm so what am i supposed to do , i can't think of any number at which it becomes zero

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0I think I know something that might work

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0let f(r) = r^4 +2r^2 +r f ' (r) = 4r^3 +4r = 4r ( r^2 +1 ) which as only one zero at r=0 , and it can be shown by the first derivative test that this is a minimum

anonymous
 5 years ago
Best ResponseYou've already chosen the best response.0so , when r=0 when have a minimum turning point , and since r=0 is a zero of the function that means the function oly has one zero ( at r=0 ) so r=0 is the only real solution
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