Question

determine as raizes destas equaçao irracional a-√3x+4=√3x-11. +1

Answer 1

$\sqrt{3x+4} = \sqrt{3x-11} +1\\\\ ( \sqrt{3x+4})^2 = (\sqrt{3x-11} +1)^2\\\\ 3x+4 = 1+3x-11+2 \sqrt{3x-11} \\\\ 3x-3x+4+10=2\sqrt{3x-11}\\\\ 14 = 2\sqrt{3x-11}\\\\ (14)^2=(2\sqrt{3x-11})^2\\\\ 196 = 4(3x-11)\\\\ 196 = 12x-44\\\\ 12x = 240\\\\ x= \frac{240}{12}\\\\ \boxed{x=20}$

Answer 2

 
 (√3x+4)² = (√3x-11 +1)²   
         
 3x + 4 = 3x-11 + 2√3x-11 + 1

 3x + 4 - 3x - 2√3x-11   = - 4 - 11 + 1
 
 - 2√3x-11 = - 14
 
    √3x-11 = -14
                   -2

(√3x-11)²  =  (7)²
    3x - 11 = 49
     3x = 49+11
      3x = 60
       x = 20