Can each root equation be rearranged to a corresponding polynomial equation? Let $n\in\mathbb{N}_+$.
Can each root equation of $n$ unknowns be rearranged to a polynomial equation of $n$ unknowns whose solution set contains the solution set of the root equation?
If not, is this true at least for root equations of one or two unknowns?
I already know the solution method with raising both sides of the root equation to the same power. But I guess this method is limited to simple root equations because raising to a power of a sum of at least three summands on one side of the equation does not reduce the number of roots on this side of the equation. Take e.g. the equation $\sqrt{A(x)}+\sqrt{B(x)}+\sqrt{C(x)}=\sqrt{D(x)}+\sqrt{E(x)}$.
I already know the method of introducing new unknowns. Does this method answer my question?

By irrational equation, I assume you meant one with one or more variable under a radical as assigned here. The technique of raising both sides to the same power may be invoked repeatedly if needed until there are no radicals. Take, for example.
$$A=\sqrt{x}+\sqrt{y}\implies A^2=\big(\sqrt{x}+\sqrt{y}\big)^2 = 2 \sqrt(x) \sqrt(y) + x + y$$
$$A^2 = 2 \sqrt(x) \sqrt(y) + x + y\implies A^2-x-y=2\sqrt{x}\sqrt{y}$$
$$\big(A^2-x-y\big)^2=\big(2\sqrt{x}\sqrt{y}\big)^2\implies
 A^4 - 2 A^2 x - 2 A^2 y + x^2 + 2 x y + y^2 = 4 x y$$
$$A^4 - 2 A^2 x - 2 A^2 y + x^2 - 2 x y + y^2 = A^4-2A^2(x+y)+(x-y)^2=0$$
The last is an algebraic equation with no radicals an no loss of the "solutions" of the original. The example has been limited to two unknowns and of the same $[(1/2)]$ power. We can, to an extent, also mix the powers of the radicals as in:
$$A=\sqrt{x}+\sqrt[3]{y}\implies (\sqrt[3]{y})^3=\big(A-\sqrt{x}\big)^3
=A^3 - 3 A^2 (x)^{1/2} + 3 A x - x^{3/2}$$
$$\implies y = A^3-3Ax-\sqrt{x}(3A^2+x)$$
$$\big(\sqrt{x}(3A^2+x)\big)^2=\big(A^3-3Ax-y\big)^2$$
$$\implies 9 A^4 x + 6 A^2 x^2 + x^3=A^6 - 6 A^4 x - 2 A^3 y + 9 A^2 x^2 + 6 A x y + y^2$$
$$\implies A^6 - 15 A^4 x - 2 A^3 y + 3 A^2 x^2 + 6 A x y - x^3 + y^2 = 0$$
