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

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Homework Help Overview

The discussion revolves around the equation 1/a^3 + a^2 + 9 = 0, which is identified as a fifth-degree polynomial after manipulation. Participants are exploring the nature of its roots, particularly the possibility of a negative real root.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • Participants discuss the transformation of the original equation into a fifth-degree form and reference DesCartes' rule of signs to analyze the roots. Questions arise regarding the implications of the signs of the polynomial at specific values of 'a'. There is also a focus on clarifying the initial equation's structure and the implications of multiplying by a^3.

Discussion Status

The discussion is ongoing, with participants providing insights into the nature of the roots and questioning the initial interpretation of the equation. Some guidance has been offered regarding the application of DesCartes' rule of signs, but there is no explicit consensus on the conclusions drawn from the analysis.

Contextual Notes

Participants note the potential confusion stemming from the initial equation's format and the implications of treating it as a cubic equation. There is an acknowledgment of the GRE context, which may impose specific constraints on the problem-solving approach.

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

This a GRE question. Thanks!
 
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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?
 
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I think you may have missed the '1/' at the beginning, Halls. that one does not make it more cubic, though.
 
jacobrhcp said:
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.
 
HallsofIvy said:
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.
 

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