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shybaby · Nov 28, 2009, 04:13 PM

n2+4n-12=0

medic-dan · Nov 28, 2009, 04:44 PM

Homework question?

Get rid of the -12 side by adding 12 to both sides.

n2+4n-12+12= 0+12

n2+4n = 12

Now can you solve it?

mathwiz3502 · Nov 28, 2009, 04:57 PM

This will help you for any of these problems
n^2+4n-12=0
factor
(n-2)(n+6)=0
now since these are multiplied the zero property of multiplication says that at least one of these must equal 0
so you get
n-2=0
and
n+6=0
solve them and you get
n=2 and -6
for proof
2^2+2(4)-12=0
4+8-12=0
0=0
(-6)^2+-6(4)-12=0
36+-24-12=0
0=0
give me feedback

medic-dan · Nov 28, 2009, 08:53 PM

I thank you mathwizard. It's been a while but now that you post that I'm thinking FOIL again.

morgaine300 · Nov 28, 2009, 11:58 PM

Quite frankly, I don't see the point of all that. Subtract the constant, combine like terms, divide, you're done.

Of course, OP didn't say what the instructions actually were, but without seeing otherwise, I assume solving for n.

mathwiz3502 · Nov 29, 2009, 05:20 AM

morgaine, many people coud not solve
n^2+4n=12 because they would not think of negative 6, and since I'm actually taking the same course, no joke in acc. int. advanced math 1 and some off 2, so I know there will soon be harder equations which is why I tell him the "long 100% way."

Unknown008 · Dec 1, 2009, 08:01 AM

There's another method, that however is a little more complex. It's called completing the square. Demonstration:

$n^2+4n-12=0$

$(n+2)^2 - 4 -12=0$

$(n+2)^2 -16=0$

$(n+2)^2 =16$

$(n+2) =\sqrt{16}$

$n+2 =\pm 4$

$n=\pm 4 - 2$

So, n = -4 - 2 = -6 or
n = 4 - 2 = 2