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I have two curves. The first is a simple parabola described by

$$y = x^2+bx+c$$

The second is a little more complicated and is described by

$$y = ((n-x^2)/2/x)^2$$

I am trying to find the $x$ and $y$ intersections of those two curves (provided they intersect) but I cannot work out the formula to use.

I tried the trustworthy WolframAlpha and what it returned seemed to be far too complex. Is it really that complex?

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  • $\begingroup$ How should we interpret the double division in the second formula? But yes, the answer is probably quite complex, as you are essentially trying to find the roots of a polynomial of degree 4 or 6 (depending on the aforementioned clarification). $\endgroup$ – Greg Martin Oct 11 '16 at 8:21
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Using the input to Wolfram Alpha $$\frac{\left(n-x^2\right)^2}{4 x^2}=x^2+b x+c$$ reduces to $$3 x^4+4 b x^3+2(2 c+n) x^2-n^2=0\tag 1$$ Now, have a look here and enjoy the fun of the analytical solutions of quartic polynomials !

You could better consider numerical methods or, using graphics, consider the intersection of the two functions $$f(x)=3x^2+4bx+2(2c+n) \qquad \qquad g(x)=\frac {n^2}{x^2}$$ This will tell you that there are exactly two real roots.

If, in the Wikipedia page, you look at the pleasant $\Delta$ and apply it to $(1)$, you should get $$\Delta=-2304 n^4 \left(8 c^2+8 c n+5 n^2\right)$$ which is always negative which implies two distinct real roots and two complex conjugate non-real roots.

Considering the behavior of functions $f(x)$ and $g(x)$, you know that one of the roots is positive while the other one is negative.

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  • $\begingroup$ Thank you. I wont profess to understand. But I now have something to study further. $\endgroup$ – DeveloperChris Oct 11 '16 at 11:08
  • $\begingroup$ @DeveloperChris. It is not so complex if you consider what are the plots of the two functions. Just to be nice to me, take a sheet of paper and a pen. Now draw the functions (you do not need to put values : the trend is more than sufficient). Cheers and good luck in your studies. $\endgroup$ – Claude Leibovici Oct 11 '16 at 18:22
  • $\begingroup$ This is me playing with cryptography (as a hobby). The smallest number I use is 128 bit and I need accurate results so graphing is not feasible. There are alternatives I can use so I am looking at that. $\endgroup$ – DeveloperChris Oct 12 '16 at 4:08
  • $\begingroup$ @DeveloperChris. I do not understand your comment. Could you provide me a few values of $b,c,n$ ? In any manner, the Wikipedia page gives the formulae for the roots. $\endgroup$ – Claude Leibovici Oct 12 '16 at 4:43
  • $\begingroup$ When I get home tonight I'll post some example values into the question. $\endgroup$ – DeveloperChris Oct 12 '16 at 5:23

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