+ 
 = 
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