So, when I was posting an answer to another user's question, I came up with this strange-looking equation whose roots I want to find. The main problem I have is that the radical sign is really throwing me off, and I'm not sure how I should deal with it. Can someone help me out? (Also, I'm not sure what tag to put this under. I don't think this technically counts as a polynomial, does it?) $$y=\frac {-8\pi x} {\sqrt {2704-4x^2}}+2\pi \sqrt {2704-4x^2}+4\pi x$$
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2$\begingroup$ The analytical expression for the roots $y(x) = 0$ are truly horrible. I would go for a numerical solution here:) $\endgroup$– WintherMar 9, 2015 at 8:21
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$\begingroup$ Note that the expression inside the radical factorizes as $(52-2x)(52+2x)$; see if that helps $\endgroup$– P VanchinathanMar 9, 2015 at 8:25
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3 Answers
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Since you only care about roots. Consider
$$\frac {-8\pi x} {\sqrt {2704-4x^2}}+2\pi \sqrt {2704-4x^2}+4\pi x=$$
$$\frac{-8\pi x+2\pi(2704-4x^2)+4\pi x\sqrt{2704-4x^2}}{\sqrt{2704-4x^2}}$$
This is 0 if the numerator is 0. So we are solving
$$-8\pi x +2\pi(2704-4x^2)+4\pi x\sqrt{2704-4x^2}=0$$
move $\sqrt{ }$ over and square to get
$$(8\pi x -2\pi(2704-4x^2))^2=16\pi^2x^2(2704-4x^2)$$
This is a degree 4 equation which you can try and solve by factoring if possible or using the awful Cardano formulas. I have left out a fair amount of equivalence checking (denominator 0 for discarding it, and positive/negative values for squaring) but that's just a little more work I think.
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You could put $y-4\pi x$ on the left-hand side, then square both sides. Any solution of your equation will also be a solution of the new equation, which has no square-roots any more.
I'm not sure what to do next.
$$(y-4\pi x)^2=\frac{64\pi^2x^2}{2704-x^2}-32\pi^2 x+4\pi^2(2704-x^2)$$
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$\begingroup$ How do you get that there are no square roots after squaring? There is a sum of squares on the right hand side no? $\endgroup$– DRFMar 9, 2015 at 8:15
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Try multiplying both sides with multiplicative inverse of $1/ ( 2704 - 4r^2 ) ^ {1/2}$, it should reduce to a quartic equation in $r$.
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$\begingroup$ More likely to be a quartic equation, possibly with spurious solutions $\endgroup$– HenryMar 9, 2015 at 8:10
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$\begingroup$ What is $r$? You should expand this answer, also using LaTeX helps. $\endgroup$– tomaszMar 9, 2015 at 8:24
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$\begingroup$ r is same as x , if i had latex i would, don't have enough mb of data to download it, i am a student... $\endgroup$– CloverrMar 9, 2015 at 8:28
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$\begingroup$ @Nilanjan: You don't have to download LaTex as this forum has a built-in feature. If you put dollar signs at each end of an expression, it transforms for example, a^2+b^2 = c^2 to $a^2+b^2 = c^2$. $\endgroup$ Mar 9, 2015 at 8:42
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$\begingroup$ ohh, that i didnt knew, now i too can edit the answers like a pro a^2$ ! Ty ty $\endgroup$– CloverrMar 9, 2015 at 8:57