√3x+1 −x+3=0
 (√3x+1)^2 =(x-3)^2
 3x + 1 = x^2 - 6x + 9
 0 = x^2 - 9x + 8
 0= (x - 8) (x – 1)
  X = 8 and x = 1
 First let’s solve for x = 8
 √3 (8) + 1 – (8) + 3 = 0
 √24 + 1 – 5 =0
 √25 – 5 =0
 5-5=0
 0=0
 X=8 is a solution
 For x=1
 √3 (1) + 1 – (1) + 3 = 0
 √4 – 4 =0
 2 – 4 =0
 -2=0
 X=1 is not a solution 
 Therefore:
 a. There is only one solution: x = 8. The solution x = 1 is an extraneous solution.