- Tue Dec 04, 2012 1:29 am
- Forum: Intermediate Algebra
- Topic: Question on deriving the quadratic formula
- Replies:
**5** - Views:
**3220**

I said where I got stuck in my original post. If you want me to write out the entire thing though. 1. ax 2 + bx + c = 0 2. ax 2 + bx = -c 3. x 2 + bx/a = -c/a 4. x 2 + bx/a + b 2 /4a 2 = -c/a + b 2 /4a 2 5. (x + b/2a) 2 = -c/a + b 2 /4a 2 This is where I'm stuck. How do you get 4a/4a from the a in -...

- Mon Dec 03, 2012 9:39 pm
- Forum: Intermediate Algebra
- Topic: Question on deriving the quadratic formula
- Replies:
**5** - Views:
**3220**

I have only one question for this. Let me write out where I got stuck. (x + b/2a) 2 = -c/a + b 2 /4a 2 What I can't figure out is how the common denominator of 4a is gotten. Let me give an example. Lets say we want to add together 1/4 and 5/8. We would look of course at the multiples of 4 and find o...

- Wed Nov 28, 2012 10:58 pm
- Forum: Advanced Algebra ("pre-calculus")
- Topic: Question about imaginary numbers
- Replies:
**2** - Views:
**3810**

I think I get it now, thanks. I get stuck on the stupidest stuff sometimes :P.

- Wed Nov 28, 2012 7:24 pm
- Forum: Advanced Algebra ("pre-calculus")
- Topic: Question about imaginary numbers
- Replies:
**2** - Views:
**3810**

Ugh. This is a stupid question, but unfortunately it's impossible for me to let this go until I know the answer. Alright, i goes in a pattern yes? i 0 = 1 i 1 = i i 2 = -1 i 3 = -i i 4 = 1 And so on. Now for the question :roll: I see a lot that the reason i 4 = 1 is as follows. (i 3 )(i 1 ) = i 4 Si...

- Wed Nov 21, 2012 1:33 am
- Forum: Intermediate Algebra
- Topic: Question on completing the square
- Replies:
**2** - Views:
**1930**

Ahhhhhh I see. I hate and love math because of tricky dicky shit like this lol. Thanks.What is (x - 1/4)x^{2}- 1/2x + 1/16 = 5/4 + 1/16.

Now we have (x - 1/4)^{2}= 21/16

My question is, what happened to the 1/16 on the left side?^{2}equal to, after expansion?

- Tue Nov 20, 2012 10:45 pm
- Forum: Intermediate Algebra
- Topic: Question on completing the square
- Replies:
**2** - Views:
**1930**

Something about the lesson given here confuses me. I understand how to complete the square, but this is bothering me and I need to know why this happens. So we have sample question 4x 2 - 2x - 5 = 0 I understand everything up until I have to convert the left side to squared form, let me show you why...

- Tue Nov 06, 2012 8:39 pm
- Forum: Beginning Algebra
- Topic: How should I treat -x^2?
- Replies:
**4** - Views:
**3377**

Ah ok. Thanks .

- Tue Nov 06, 2012 2:50 am
- Forum: Beginning Algebra
- Topic: How should I treat -x^2?
- Replies:
**4** - Views:
**3377**

Wouldn't -x^{2} mean -x * -x which would result in a positive? Compared to -(x^{2}), for the sake of this lets say x is 2. So -(2^{2}) = -(4) = -4.

- Tue Nov 06, 2012 2:00 am
- Forum: Beginning Algebra
- Topic: How should I treat -x^2?
- Replies:
**4** - Views:
**3377**

From what I understand there's a difference of opinions when it comes to this. Some people look at it as -(x^2) and others -x^2. Both give different answers which is why I'm asking what should I do when I have something like in front of me?

- Thu Oct 25, 2012 11:08 am
- Forum: Intermediate Algebra
- Topic: Question about sum and difference of cubes
- Replies:
**2** - Views:
**1804**

I think I figured it out on my own. x^{3}y^{6} becomes (xy^{2})^{3} because the three is getting factored out right? I hope that's right :l.