Solving an Odd Equation: Two Different x Values for Same F?

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The discussion revolves around solving the equation F(x) = Ax / (bx² + c), which is an odd function yielding the same positive and negative solutions for each F. The user encounters two different x values for the same F, leading to confusion about the nature of the solutions. It is clarified that the equation can be transformed into a quadratic form, resulting in two possible x values for given constants a, b, and c. The concern about obtaining ± the same number as roots is acknowledged, and the user is encouraged to share their working for further assistance. The conversation emphasizes the nature of quadratic equations and their solutions in relation to the original function.
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Hello everyone
I want to solve this equation
F(x)=Ax/((bx*x)+c =constant which is an odd function in which we get same positive and negative solutions for each F
But If I solve his equation I get two different x values for The same F any suggestions??
Thank you
 
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eahaidar said:
Hello everyone
I want to solve this equation
F(x)=Ax/((bx*x)+c =constant which is an odd function in which we get same positive and negative solutions for each F
But If I solve his equation I get two different x values for The same F any suggestions??
Do you mean two different x values for the same value of what you're calling the constant?

Graph y = x/(x^2 + 1) to get an idea of what your function F looks like.
 
First thank you for the reply
Second yes these are all constants but what I am asking about is how come if I solve it analytically I would not get 2 values on x same but opposite to each other
But different values of x
 
yes these are all constants
... surely you are treating x as a variable?
You seem to be wanting to solve: $$a=\frac{x}{bx^2+c}$$... here a=<constant>/A which is also a constant.

This becomes ##abx^2-x+ac = 0## which is a quadratic equation.

You are getting two possible values of x for given values of a,b, and c because there are two possible values that make the relation true. What is the problem?

Your concern seems to be that you are getting ##\pm## <the same number> as the roots ... if so, then please show your working.
 
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