2
$\begingroup$

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?) $$y=\frac {-8\pi x} {\sqrt {2704-4x^2}}+2\pi \sqrt {2704-4x^2}+4\pi x$$

$\endgroup$
  • 2
    $\begingroup$ The analytical expression for the roots $y(x) = 0$ are truly horrible. I would go for a numerical solution here:) $\endgroup$ – Winther Mar 9 '15 at 8:21
  • $\begingroup$ Note that the expression inside the radical factorizes as $(52-2x)(52+2x)$; see if that helps $\endgroup$ – P Vanchinathan Mar 9 '15 at 8:25
3
$\begingroup$

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.

| cite | improve this answer | |
$\endgroup$
0
$\begingroup$

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.
$$(y-4\pi x)^2=\frac{64\pi^2x^2}{2704-x^2}-32\pi^2 x+4\pi^2(2704-x^2)$$

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ How do you get that there are no square roots after squaring? There is a sum of squares on the right hand side no? $\endgroup$ – DRF Mar 9 '15 at 8:15
  • $\begingroup$ It is the same square-root in each term. $\endgroup$ – Empy2 Mar 9 '15 at 8:17
  • $\begingroup$ Ahh good point.:) $\endgroup$ – DRF Mar 9 '15 at 8:19
0
$\begingroup$

Try multiplying both sides with multiplicative inverse of $1/ ( 2704 - 4r^2 ) ^ {1/2}$, it should reduce to a quartic equation in $r$.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ More likely to be a quartic equation, possibly with spurious solutions $\endgroup$ – Henry Mar 9 '15 at 8:10
  • $\begingroup$ What is $r$? You should expand this answer, also using LaTeX helps. $\endgroup$ – tomasz Mar 9 '15 at 8:24
  • $\begingroup$ 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... $\endgroup$ – Cloverr Mar 9 '15 at 8:28
  • $\begingroup$ @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 $a^2+b^2 = c^2$. $\endgroup$ – Tito Piezas III Mar 9 '15 at 8:42
  • $\begingroup$ ohh, that i didnt knew, now i too can edit the answers like a pro a^2$ ! Ty ty $\endgroup$ – Cloverr Mar 9 '15 at 8:57

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.