anonymous
  • anonymous
r^4 + 2r^2 + r = 0 ... How to solve this ???
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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chestercat
  • chestercat
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anonymous
  • anonymous
factor out r
anonymous
  • anonymous
r (r^3 +2r +1 )
anonymous
  • anonymous
for the second term use factor theorm

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anonymous
  • anonymous
note, if you put r=-1 in the second bracket then it is zeroed out so (r+1) is a factor
anonymous
  • anonymous
therefore (r^3 +2r +1 ) = (r+1)(r^2 +ar ) let r=1 4 = 2(1+a) 1+a = 2 a= 1
anonymous
  • anonymous
umm if i put r = -1 in the second bracket the answer doesn't come out to be zero :S
anonymous
  • anonymous
ohh yeh, made mistake there
anonymous
  • anonymous
umm so what am i supposed to do , i can't think of any number at which it becomes zero
anonymous
  • anonymous
I think I know something that might work
anonymous
  • anonymous
let 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
  • anonymous
so , 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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