We wonder if each of the radical expressions on the left equals to one of 
the radical expressions on the right.

We try the first expression on left equaling to the first
expression on the right






We check to see if that makes the other radicals equal:






No, those are not equal.  So 2 is not a solution.

So we try the first expression on the
left equaling to the second expression on the right:



 



We could solve this by the quadratic formula but there 
would be no use unless also the second expression on the left
equals the first expression on the right.  So we'll wait and 
see if that can be the case:

To see if that can be the case, we set those equal:



Square both sides. The square of a fourth root equals
the square root:



Square both sides again:





We're in luck! That just happens to be the same 
quadratic, so we will solve it by the quadratic formula:



















We must check solutions of equations containing radicals,
as they often have extraneous answers.

All the roots in the equation have even indices, so the
four radicands must not be negative. 

We only need to make sure that there are no negative 
numbers under any of the radicals:

The first radicand is



Substituting 

, that's positive.  So far so good.

Substituting 

, that's positive also.  OK.

The second radicand is



Substituting 

, that's positive

Substituting 

, that's positive also.

The third radicand is



Substituting 

, that's positive

Substituting 





Oh oh, that's negative, so 
is an extraneous answer and must be discarded.

Thus the only solution possible is , but we
must make sure it gives a positive radicand in the last 
expression on the right:

The fourth radicand is



Substituting 







, that's positive.

So there is one solution:



Edwin