<s>[INST] How to remove roots from an equation? The question here is how (if it is even possible) to remove the square root terms and transform the following equation to a polynomial with one unknown $x$.
The coefficients $a$, $b$, $c$, and $d$ are known and also $r$.
$$a \sqrt{x} + b \sqrt{x} \sqrt{r^2-x^2} + c \sqrt{r^2-x^2} + d = 0$$
 [/INST] Well the $r-x^2$ screams that they want a trig substitution as lone student's answer suggest.
But you can always remove roots but bringing terms over and squaring.
$a \sqrt{x} + b \sqrt{x} \sqrt{r^2-x^2} + c \sqrt{r^2-x^2} + d = 0$
$\sqrt{x}(a + b\sqrt{r^2-x^2}) = - d - c\sqrt{r^2-x^2}$
$x(a^2 + b(r^2-x^2) + 2ab\sqrt{r^2-x^2})) = d^2 + c^2(r^2 - x^2) - 2cd\sqrt{r^2-x^2}$
$x(a^2 + b(r^2-x^2))- d^2 - c^2(r^2-x^2) = -(2cd+2abx)\sqrt{r^2 -x^2}$
$(x(a^2 + b(r^2-x^2))+ d^2 + c^2(r^2-x^2))^2 = (2cd+2abx)^2(r^2-x^2)$ so
$(x(a^2 + b(r^2-x^2))+ d^2 + c^2(r^2-x^2))^2 -(2cd+2abx)^2(r^2-x^2) =0$
Which is a $6$th degree polynomial
which 1) Answers exactly what you asked and 2) makes things much, much, much worse.
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