Hello there!
Finding the solution will require us to write this down in numbers. Then after that, we will solve...
sqrt(x+2)+4=6
First let's move the 4 to the other side...
sqrt(x+2)=2
Now we need to square both sides so that we can get x out of the radical...
[sqrt(x+2)]²=2²
x+2=4
x=2
Now is this solution extraneous? We need to plug this BACK INTO the original equation and if we get a negative under a radical, we know it's extraneous. Also, if
sqrt(x+2)+4=6
sqrt(2+2)+4=6
sqrt(4)+4=6
2+4=6
Well it looks like we are correct! 2 is NOT an extraneous solution!
I hope this helps!
Best wishes :)
