Let the answer be A: 



Then square both sides:



Let 



Then 

Then square both sides of the equation for B:



Then 

So we have the system of equations



Solve the system by substitution. [If you have trouble solving
that system, tell me in the message form below and I'll get back
to you by email.]

You get two real solutions and two imaginary solutions. However, 
since by definition, square root radicals always indicate only
non-negative real numbers, both A and B are non-negative real
numbers.  So the only acceptable solution to the given problem 
is 

A = 2  (B = 5/3).


Edwin

    Evaluate    

Let  us consider, for brewity of writing, more general expression

 = x,    (1)

where  a = ,  b = .  Square (1)  (both sides).  You will get then  

 = ,

 = .        (2)


Square (2)   (both sides).  You will get then

 = .         (3)

 = .         (4)


Notice that the right side of the expression (4) is the same as the given expression,  so it is equal to x.  Thus you have 

 = x.                  


It is equivalent to


 -  = x,    or

 = 0.


Now substitute here  a = ,  b = . You will get this equation in the form

 = 0,   or, multiplying all the terms by 3

 = 0.


The plot of the last polynomial is shown below.






Plot y = 



It clearly shows that  x= 1  and  x= 2 are the roots.  Having this HINT, you can check it MANUALLY  (as I did . . . ).

The two other roots of the polynomial are complex numbers.


Since  the value of    is, obviously, real number greater than 1, it can be only 2.


It proves that   = 2.


Answer.   = 2.


Check.    = 1.984 (approximately).