How to solve what should be a simple variant of a quadratic

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Justin R
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Hi everyone,

For some reason I can't put this together in my head. I have a series of equations that can be solved directly, but my resulting equation is of the form:

Ax^2 + Bx + Cx^(-1) + D = 0

Obviously the C/x term is the one throwing the wrench in the works here. Could anyone point me in the right direction?

Thanks in advance!

-Justin
 
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Justin R said:
Hi everyone,

For some reason I can't put this together in my head. I have a series of equations that can be solved directly, but my resulting equation is of the form:

Ax^2 + Bx + Cx^(-1) + D = 0

Obviously the C/x term is the one throwing the wrench in the works here. Could anyone point me in the right direction?

Thanks in advance!

-Justin

Nevermind... It's very easy. Turn it into a cubic equation by multiplying through by X, resulting in the readily solvable:

Ax^3 + Bx^2 + Dx + C = 0
 
And be sure to note that while x= 0 might satisfy the third degree equation, it cannot be a solution to the first equation.