- Thread starter ryloth
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Are you SURE it's "Quadratic"? It certainly look's quadratic in the denominators, but does that make the whole thing "quadratic"?I am trying to solve the attached quadratic parametric equation.

I have just simplify a bit the denominator by collecting a couple of terms, but I am stuck there.

Thank you a lot.

View attachment 11006

Sorry for the low quality of the image.

Gianluca

What do you mean by "collect"? Factoring may be a nice direction.

\(\displaystyle \dfrac{1}{(x+2k)(x-(k+1))} = \dfrac{x}{(x+2k)(x-2k)}\)

A common denominator may help you on your way.

Note: "Parametric" usually means that there is some independent variable that controls all the other variables. It doesn't normally mean that there is a arbitrary parameter included ('k' in your example). Anyway, your question was understandable.

Indeed I am not used to math terms (I am more confident with biochemistry) and vocabulary gives not much help.

In italian those kind of equations are defined as 'equazioni di secondo ordine parametriche'.

I was eventually able to solve it.

Gianluca

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\(\displaystyle \displaystyle{\dfrac{1}{x^2 + (k-1)x - 2k^2 - 2k} = \dfrac{x}{x^2 - 4k^2}}\)I am trying to solve the attached quadratic parametric equation.

I have just simplify a bit the denominator by collecting a couple of terms, but I am stuck there.

Thank you a lot.

View attachment 11006

Sorry for the low quality of the image.

Gianluca

Last edited:

- Joined
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Fair enough. Thanks for sharing the translation.

Indeed I am not used to math terms (I am more confident with biochemistry) and vocabulary gives not much help.

In italian those kind of equations are defined as 'equazioni di secondo ordine parametriche'.

I was eventually able to solve it.

Gianluca