How to remove roots from an equation?

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.

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.