Whenever you have to square both sides to solve, you have to check for extraneous solutions.

That tells you *m* could be 2 or –10, but because part of the solution was squaring both sides, you need to run both possible solutions through the original equation.Try 2 first:

That works, now how about –10?

Nope. Remember that the square root function √ returns only positive results, so –10 is an extraneous solution that doesn’t work in the original equation. The sum of all solutions is just 2.

One more note here: I often think it’s worth graphing questions like this.

It should be obvious from that graph that there’s only one intersection of the two functions, at *x* = 2.

from Tumblr https://ift.tt/2B5QuHh