
Squre root of 1 I guess
I begin with $\displaystyle x^2+8x=15$. Since they say to solve by completing the square, I ignore that it can be factored as is and add one to each side. Since $\displaystyle (1/2*8)^2=16$. And get $\displaystyle x^2+8x+16=1$ after moving the 15 and adding one to each side.
First off have I done everything correctly? Keep in mind that the problem asks for me to complete the square, which I think I have done.
Secondly is the resulting factored equation just asking me for the square root of one? $\displaystyle (x+4)^2=1$ is the same as $\displaystyle a^2=1$. Since x would be what ever number plus four squared is equal to one.
Just from factoring out the initial equation you can tell the answers are 3 and 5. So something has too be wrong.

When solving (rather than merely simplifying), you have to account for all possible values. So while $\displaystyle \sqrt{1}\, =\, 1$ for simplifying, you need to use $\displaystyle \sqrt{1}\, =\, \pm 1$ for solving.
So you actually have:
$\displaystyle x\, +\, 4\, =\, \pm 1$
Subtract the 4 over to the righthand side, and simplify to get the two values. They should match what you've been given. (Wink)

OK thats why, I'm just forgeting that the square root of 1 is 1 and didnt work the problem out all the way.
Thanks