Question 249418
sqrt(3x^2-8)+x=0
sqrt(3x^2-8)=-x
square both sides
3x^2-8=x^2
subtract x^2 from both sides

2x^2-8=0
2x^2=8
x^2=4
x=+\-2
check in original
sqrt(3x^2-8)+x=0
sqrt(3*2^2-8)+2
sqrt(3*4-8)+2=0
sqrt(12-8)+2=0
sqrt(4)+2=0
+\-2+2=0
-2+2=0 ok
+2+2=0
4=0 not ok
We get a similar result with -2
sqrt(4)-2=0
2-2=0 ok
-2-2=0 
-4=0 not ok

sometimes it works and sometimes it doesn't