Simple Question

[Zfuss12]

The solution set of radical (x+6)=x is:


I squared both sides and got x+6=x^2
then i got the quadratic x^2-x-6=0
which yields the resuls x=-2 and x=3

I think -2 rejects but I'm not sure, can somebody please reitterate and just help me to confirm.

 

[tkhunny]

Look at your original problem statement. There are two important chunks of information looking back at you.

Most obvious, x+6 > 0 ==> x > -6, since you probably want solutions in the Real Numbers.

Next, that radical-thing mean the "Principal Square Root." In the Real Numbers, that's the positive one. This means necessarily that x > 0.

The second consideration eliminates x = -2 as a solution.

Of course, you can just check your answers and come to the same conclusion.