Question

Given that a 14 years ago

Answer 1

how bout -c < x < -a

Answer 2

that make sense?

Answer 3

not really..O.o

Answer 4

got
\[\frac{(x+a)(x^2+2bx+b^2)}{x+c}<0\]

Answer 5

i thought it as x<a<c

Answer 6

well you are solving for x yes? so what you wrote says x < a

Answer 7

here is my reasoning, it may be wrong

Answer 8

\[\frac{(x+a)(x+b)^2}{x+c}<o\]
and the perfect square part is always non - negative so you are left with
\[\frac{x+a}{x+c}<0\] if \[x\neq{-b}\]

Answer 9

Idk... i thought (x+a)*(X+b)^2/(x+c)
so it depends from x+a and X+c....
x+a<0--->x<-a
x+c<0--->X<-c u were right..i 4got minuses

Answer 10

ok whew. so x is between -c (which is negative) and -a (which is positive)

Answer 11

to be complete you should also say that
\[x\neq-b\] because it would be 0 there

Answer 12

yea...Thnx a loot ^_^

Answer 13

welcome!