If 1/a^3+ a^2+ 9=0 , is ''a'' greater than/less than or equal to -1/3.
This a GRE question. Thanks!
That doesn't look like a cubic equation to me. If you multiply both sides by a3 you get a5+ 9a3+ 1= 0, a fifth degee equation. We can tell by "DesCartes' rule of signs" that it has no positive real root and only one negative real root. When a= 0, (0)5+ 9(0)+ 1= 1 which is positive and when a= -1/3, (-1/3)5+ 9(-1/3)+ 1= -2+ 1/243 which is negative. What does that tell you?
I think you may have missed the '1/' at the beginning, Halls. that one does not make it more cubic, though.
He actually multiplied by a^3, but wrote a^2.
Thanks, to solve it quickly i thought it as cubic overlooking the fact that it is actually a^-3, not a^3.
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