Verifying roots of radical function Suppose you have a consistent radical function $R(x)$ that can be solved by finding the roots of an nth degree polynomial, assumed to be solvable using radicals. Some of the $n$ roots maybe extraneous. Typically the student is told to verify each possible root by inserting it into $R(x)$ and verifying that $R(x_i) = 0$ for $i= 1,..,n$. This step basically amounts to verifying an identity that can involve multiple roots and powers. And the $x_i$'s can be complex. Can it be proved algebraically that each extraneous root can be identified, i.e., $R(x_i) \neq 0$ for some $i$?
Let me try to be specific. Suppose $R(x) = \sqrt{x+a} -(x+b) = 0$. The solution to this simple radical equation results in a quadratic equation. When two roots result it appears that one of them will be extraneous. The roots can be complex involving irrational values. Verifying that a root of the polynomial is also a root of $R(x)$ requires proving an identity: $R(x_i)=0$. This could be very difficult because the root may involve nested radicals and $R(x)$ can result in more nested radicals. But how does one prove the identity without resorting to steps that were used to generate the polynomial? I hope this helps. My gut feeling is that one cannot be assured of proving the identity.
 A: $\root n\of{a+bi}$ can always be evaluated by going to polar form. By going back and forth between polar and rectangular form, any expression $R(x)$ can be evaluated where $R$ and $x$ are of the form envisaged in the question. Then you can tell whether or not it's zero. In principal. In practice, I can see where there may be some difficulties with this approach. First of all, you get $n$ different values for $\root n\of{a+bi}$, and I guess you have to test all of them, and keep some consistency in branches chosen. Also, expressions can get unwieldy, unless you go to floating point - and then if you get zero, you don't know if it's really zero or just machine zero (and if you don't get zero, maybe that's round-off error on something that really is zero). 
A: If R(x) is a polynomial in a field F and $\alpha$ involves only a finite number of nested root extractions you can compute the minimal polynomial of $\alpha$ in F and compare it to R(x). Call the minimal polynomial of $\alpha$ g(x). Since the minimal polynomial is the unique polynomial of smallest degree in F so that g($\alpha$)=0, g(x) is a factor of R(x) iff R($\alpha$)=0.
