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 $r$.

$$a \sqrt{x} + b \sqrt{x} \sqrt{r^2-x^2} + c \sqrt{r^2-x^2} + d = 0$$

  • $\begingroup$ See Rationalizing radicals. $\endgroup$ – Dave L. Renfro Feb 26 at 21:14
  • $\begingroup$ As @DaveLRenfro alludes, removing roots is only possible if we can exactly factor the expression which is set to equal zero. $\endgroup$ – hardmath Feb 26 at 21:28

Yes it is possible. Take $\color{red}{x=r \cos t, r>0}$. Because $r^2=(-r)^2$. So, we can accept $r>0$ . Then we have,

$$a \sqrt {r}\sqrt {\cos t}+br\sqrt {\cos t}|\sin t|+cr|\sin t|+d=0$$

$$\sqrt {\cos t}\left(a\sqrt r+br |\sin t|\right)=-d-cr|\sin t|$$

$$\cos^2 t\left (a\sqrt r+br |\sin t|\right)^4=\left(-d-cr|\sin t|\right)^4$$

$$(1-|\sin t|^2)\left(a\sqrt r+br |\sin t|\right)^4=\left(d+cr|\sin t|\right)^4$$

Then $\color{red}{|\sin t|=y}$, you get

$$(1-y^2) (a\sqrt r+bry)^4-(d+cr y)^4=0.$$

Finally, you get $6$ degree polynomial respect to $y.$

I believe you can take from here.

  • $\begingroup$ @Bashar you are welcome. :) $\endgroup$ – lone student Feb 26 at 22:49

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.

  • $\begingroup$ Make things much worse? $\endgroup$ – Some Guy Feb 26 at 22:30
  • $\begingroup$ I guess he means "now the expression is a mess because it has a lot more terms. It is, indeed. Polynomial things are not necessarily easier to solve! $\endgroup$ – Andrea Marino Feb 26 at 22:35
  • $\begingroup$ Seems obviously worse to me. Admittedly it only has $6$ terms when we are done simplying an having coefficients expressed in terms of $a,b,c,d,r$ but we are no closer to solving and $\frac 23$ of our solutions will be extraneous. $\endgroup$ – fleablood Feb 26 at 22:44
  • $\begingroup$ Where you have the coefficient $b$ multiplying $(r^2-x^2)$ after the first squaring, it should be a $b^2$. Also, the $-2cd$ should be $+2cd$ $\endgroup$ – Barry Cipra Feb 26 at 23:19

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.