Step-by-step explanation:
So you can check for extraneous solutions by plugging in the 5 and -2 as x.
Plug 5 in as x
[tex]5=\sqrt{3(5)+10}\\5=\sqrt{15+10}\\5=\sqrt{25}\\5=5[/tex]
I'm not sure if this is exactly considered an extraneous solution, but I'm assuming it is, since technically the radical doesn't have a plus/minus in front of it, so it's the principle square root which means it only outputs the positive values. This would mean that the -2 wouldn't be a solution, since the principle square root doesn't output negative values, so it's a function, since if it outputted 2 values, then it wouldn't be a function by definition.
