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Jan 21 2014 11:50pm
Quote (carteblanche @ Jan 22 2014 12:46am)
take a look at your very first step:
2/( x^2 + x -12) = 0

this implies G(x) = 0
but if G(x) = 0, then G(x) exists for that value of x. if it exists, then it's in the domain. so you're already off to the wrong start. you need to find when G(x) does NOT exist.

and as for factoring:
x^2 + x -12 = 0
a = 1
b = -12
find factors of b that add to get a


What does that even mean.. Factors of b ? 1, 2, 3, 4, 6, 12..... All negative.. Now factors of b that add to get a... None of them add to get a...

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Jan 21 2014 11:52pm
Quote (fingerling @ Jan 22 2014 12:50am)
What does that even mean.. Factors of b ? 1, 2, 3, 4, 6, 12..... All negative.. Now factors of b that add to get a... None of them add to get a...


Bold'd is wrong. tell me, when does -1 * -12 = -12? Try again.
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Jan 21 2014 11:56pm
Quote (carteblanche @ Jan 22 2014 12:52am)
Bold'd is wrong. tell me, when does -1 * -12 =  -12? Try again.


So its those numbers all positive and all negative ?

-2 + 3


I guess I'm confused when u say factor.. Factoring I'm familiar with is format like

(x-2)^2

(X-2)(x-2). Etc

This is a different kind of factoring ?


So we use x^2 + ax+ b and find b factor values that add to a.. And if b is negative it is all negative and positive factors.

Now what do I do once I have -2 and 3.


Am I retarded ? This stuff is supposed to be very easy...

This post was edited by fingerling on Jan 22 2014 12:00am
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Jan 22 2014 12:06am
Quote (fingerling @ Jan 22 2014 12:56am)
So its those numbers all positive and all negative ?

-2 + 3


I guess I'm confused when u say factor.. Factoring I'm familiar with is format like

(x-2)^2

(X-2)(x-2). Etc

This is a different kind of factoring ?


So we use x^2 + ax+ b and find b factor values that add to a.. And if b is negative it is all negative and positive factors.

Now what do I do once I have -2 and 3.


Am I retarded ? This stuff is supposed to be very easy...


1) does -2 * 3 = -12?
2) yes that is factoring. x^2 - 4x + 4 factors into (x-2)(x-2). notice that the factors of 4 (-2*-2) add to get -4. so what does x^2 +x - 12 factor into?

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Jan 22 2014 12:07am
( x^2 + x -12) = 0 can be solved by factoring

or by using quadratic formula

those would be the only 2 methods you need to worry about.

what's the problem, you don't know how to factor? don't know about quadratic formula? , didn't read other posts
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Jan 22 2014 12:07am
Quote (fingerling @ Jan 21 2014 09:53pm)
the domain of G(x)= 2/ [x^2 + x - 12]
fixed^^^

To restate what the other posters have said:

The domain is every possible value x can have that won't make that function "blow up in your face".

The only way for something to go wrong with that function would be if you evaulated (plugged in a value for x) the denominator and got 0

So the domain is all real numbers except where:

x^2 + x - 12 = 0

So you need to solve that for, and it will give you value(s) for x that will make the function blow up in your face.

You can use many methods to solve it, but I will factor it:


x^2 + x - 12 = 0

(x+4)*(x-3) = 0

x = -4, x = 3

So the domain of x is all real values EXCEPT x = -4 and x = 3
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Jan 22 2014 12:12am
If you don't understand how we factored that quadratic, you need to review this material:



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Jan 22 2014 12:12am
Quote (carteblanche @ Jan 22 2014 01:06am)
1) does -2 * 3 = -12?
2) yes that is factoring.  x^2 - 4x + 4 factors into (x-2)(x-2). notice that the factors of 4 (-2*-2) add to get -4. so what does x^2 +x - 12 factor into?


I thought you said factors of b that add to a... And a is 1...

8x+x-12 is what it equals.
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Jan 22 2014 12:13am
And maybe this:

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Jan 22 2014 12:16am
Quote (fingerling @ Jan 22 2014 01:12am)
I thought you said factors of b that add to a... And a is 1...

8x+x-12 is what it equals.


you're telling me that 4 - 3 is not equal to 1?

/edit: oh, when i said factors of -12, i assumed you knew they had to multiply to get -12

-1 + 12 = 11
1 + -12 = -11
2 + -6 = -4
-2 + 6 = 4
3 + -4 = -1
-3 + 4 = 1 ---> our magic answer

This post was edited by carteblanche on Jan 22 2014 12:18am
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