+  = 

To make things easier, let the radicals be letters and only
replace them when they get squared: 

let A =  , B = , C = 

then A² = 5x+1, B² = 3x+4, C² = 16x+9  

Since A and B are on the same side they will be multiplied, so

(AB)² = (5x+1)(3x+4) = 15x² + 23x + 4 

The original problem becomes
                     
                A + B = C 

Square both sides:

             (A + B)² = C²

        A² + 2AB + B² = 16x + 9

5x + 1 + 2AB + 3x + 4 = 16x + 9

         8x + 5 + 2AB = 16x + 9

                  2AB = 8x + 4

                   AB = 4x + 2
     
                (AB)² = (4x + 2)²
 
       15x² + 23x + 4 = 16x² + 16x + 4  

                    0 = x² - 7x            

                    0 = x(x - 7)

                        x = 0;  x - 7 = 0
                                    x = 7

We must check both 0 and 7 for irrational solutions:                         

Checking x = 0

     +  = 
  +  = 
       +  = 
           +  = 
            1 + 2 = 3
                3 = 3

So x = 0 is a solution 

Checking x = 7

     +  = 
  +  = 
     +  = 
         +  = 
            6 + 5 = 11
               11 = 11

So x = 7 is also a solution

So x = 0 and x = 7 are the solutions.   

Edwin