How do you solve #(x + 4)^{2} = 245#?

1 Answer
Apr 11, 2018

There are plenty of ways to do this but I think the easiest is just to square root it. I got #x=-19.65248, 11.6525

Explanation:

The best way to do this, in my opinion, is to just square root both sides first to get ride of the square:
#√(x+4)^2 = √245#
Now you might not know but when you square root a square it cancels it out and keeps everything inside so the result is:
#x +4 =+- 15.65247584#
The next thing to do is subtract the #4# from both sides:
#x=-4+-15.65247584#
If you do not know #+-# just means you are going to get two answers which include you adding #4# and subtracting it. You have to do this every time you take the square root of something:
#x=-4+15.65247584# and
#x=-4-15.65247584#
These come out to:
#x=-19.65248# and
#x=11.6525#
You can check these by putting it back into the problem in the #x# spot as so:
#((-19.65248)+4)^2=245# and
#((11.6525)+4)^2=245#
I checked both of these and they both come out to about #245# so they you go.