Question

r^4 + 2r^2 + r = 0 ... How to solve this ???

Answer 1

factor out r

Answer 2

r (r^3 +2r +1 )

Answer 3

for the second term use factor theorm

Answer 4

note, if you put r=-1 in the second bracket then it is zeroed out

so (r+1) is a factor

Answer 5

therefore

(r^3 +2r +1 ) = (r+1)(r^2 +ar )

let r=1

4 = 2(1+a)

1+a = 2
a= 1

Answer 6

umm if i put r = -1 in the second bracket the answer doesn't come out to be zero :S

Answer 7

ohh yeh, made mistake there

Answer 8

umm so what am i supposed to do , i can't think of any number at which it becomes zero

Answer 9

I think I know something that might work

Answer 10

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

Answer 11

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