Is a Negative Real Root Possible for a Fifth Degree Equation?

[kkaman]

If 1/a^3+ a^2+ 9=0 , is ''a'' greater than/less than or equal to -1/3.

This a GRE question. Thanks!

 

[HallsofIvy]

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?

 

[jacobrhcp]

I think you may have missed the '1/' at the beginning, Halls. that one does not make it more cubic, though.

 

[Mark44]

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.

 

[kkaman]

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?

Thanks, to solve it quickly i thought it as cubic overlooking the fact that it is actually a^-3, not a^3.