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I am working on a math puzzle that results in the answer setting up a pair of equations for corresponding sides of similar triangles, then solving the first for y and substituting in the second that gives an equation with a single unknown, like this:

$$x - \frac {x} {\sqrt {16 - x^2}} - \frac {x} {\sqrt {9 - x^2}} = 0$$

Now the trick is to solve for $x$. But it has been pointed out to me that this equation is a quartic. OK, there are lots of places on the 'net that can solve the roots of quartics, no problem, as computers have come a long way. But how do I, when given a polynomial of this type, deduce that it is a cubic or a quartic or even a quintic, and solve for it but not have the coefficients of the general form? Since this equation has no $x^4$th term in it, how do I know that I'm dealing with a quartic? How can I manipulate this equation to get to the general form of $ax^4 + bx^3 + cx^2 +dx + e = 0$, thus having numbers for $a, b, c, d$, and $e$? (My TI-89 using nSOLVE gives the answer as $\pm 2.60328775442$, and $0$ for $x$, thus giving me only 3 solutions, not 4, making me think that it is a cubic)

If anyone would like the complete puzzle to see what I am working on, please ask, I am happy to supply the puzzle!

Thanks for any help!!

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    $x=0$ is a solution. Setting this possibility aside and dividing through by $x$ you have two square roots to eliminate, and doing this gives you (at most) a quartic. You can see that you will get an even function (if $x$ is a solution then so is $-x$) but your solutions may not all be real. You will also have to check that the solutions of the quartic do solve the original equation, as squaring can add additional solutions, which would be picked out by choosing a negative opposite sign for the square root.2012-06-22
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    I don't have time to write a post but it seems that there is only one real solution: the trivial one $x=0$. All the others are complex and if my calculation are correct you reach a quartic after substituting $x^2$. To visualize solutions you could also plot your equation.2012-06-22
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    I'd love to see the original puzzle, certainly!2012-06-22
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    @rubik, after dividing through by $x$ the left side is positive for $x=0$ and negative for $x$ just less than 3, so surely there's a real solution between 0 and 3.2012-06-22
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    Oh my bad sorry. I had copied the problem when it was still non-latex and I copied it wrong! I had $x$ instead $x^2$ inside the square roots...2012-06-22

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