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Old 02-13-2005, 02:20 PM   #1
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You ever think about creating a Homework Help Topic? For people who are stuck with specific problems and that may need help on how to do them when their teacher is a crazy pscho who doesn't teach very well?

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[ February 24, 2005, 04:44 PM: Message edited by: Bad Andi ]
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Old 02-13-2005, 02:48 PM   #2
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...That's a pretty good idea, actually. [img]smile.gif[/img]

This topic could suffice for that for a bit, then if we get more people needing help in different subjects, I'll just make seperate topics concerning those subjects.

Sounds like someone needs help...?
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Old 02-13-2005, 04:04 PM   #3
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Nah, I'm fine, I just thought it was a good idea.

But I want you to try this sample problem.

Find the other roots of this equation

x^3+2x^2+6x+3; 2i

Solve Andi
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Old 02-13-2005, 05:34 PM   #4
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ACK! YOU KNOW MY ONLY WEAKNESS! NOOOOO...

Errr...

x(x^2+2x+9)

x(x+ )(x+ )

...Umm...

...

...I'll appoint someone else to be the math guru person.
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Old 02-13-2005, 06:00 PM   #5
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Anyone read the death of Julius Caesar? I need help on my guided reading question.
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Old 02-13-2005, 06:44 PM   #6
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As in the Shakespearean play?

No, but I'm actually going to be reading it pretty soon, believe it or not. [img]tongue.gif[/img]
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Old 02-13-2005, 07:24 PM   #7
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I have......but probably already forgetten most of it.

Might be able to help ya.
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Old 02-13-2005, 07:34 PM   #8
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^Sorry, haven't read that one yet.

I saw the math question and I had to try!

The other roots are -2i and -0.579507... I cheated on the second one by using my calculator, but trust me, I've been trying to figure out how to get that last one without it for 20 minutes. Using Descartes' Rule of Signs and synthetic division can only get you so far... =P Yay for ICA class! (Seriously, is there a way to get that fraction without a calculator?? Now I want to know!)
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Old 02-13-2005, 07:52 PM   #9
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I need to know why in the middle of the battle that Brutus and Cassius were fighting about. I'm up to the part where Brutus's wife dies, if that can help you a bit.
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Old 02-13-2005, 08:07 PM   #10
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Quote:
Originally posted by Katyrna:
^Sorry, haven't read that one yet.

I saw the math question and I had to try!

The other roots are -2i and -0.579507... I cheated on the second one by using my calculator, but trust me, I've been trying to figure out how to get that last one without it for 20 minutes. Using Descartes' Rule of Signs and synthetic division can only get you so far... =P Yay for ICA class! (Seriously, is there a way to get that fraction without a calculator?? Now I want to know!)
Use a graphing calculator
Plug the equation in "Y=" then hit 2nd Graph, and it'll give you a table, any X where Y = 0 is a root.
And leave it at an irrational root

But I don't have my calc so I can't do this one.
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Old 02-13-2005, 08:30 PM   #11
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^ But, does that work when the answer's imaginary? Also, I was taught another way, where you can graph the equation and then look for where it intersects the x-axis. Then just go to 2nd, Trace, and then 2: zero, if you're using a TI-83 or other related model. Move the cursor to the left of where the intersection is and hit enter, and then right of the intersection and hit enter again. Put it one more time somewhere in between the two and hit enter once more--that'll show you the correct x-value.

Hint: For anybody else out there working with imaginary numbers, if you're given one of the imaginaries, the opposite of it will also be a solution. They always come in pairs as a rule. So since 2i was given, you automatically assume -2i is true, too. Wow, I finally understand enough math to answer something... O.O
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Old 02-13-2005, 08:51 PM   #12
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And, the other root is either -1 or -3
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Old 02-13-2005, 08:56 PM   #13
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When you do the PNI chart you get two possible solutions:
P N I
0 3 0
0 1 2

Since 2i Is a root, and there can never be just ONE imaginary root, but like Kat said, -2i is your other root.

Multiplying (X-2i)(X+2i) gets you X^2+4

But usually you'll get planned problems which will come out with sensible roots, and by sensible I mean solvable because 95% of math past Algebra I is unneeded crap.
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Old 02-13-2005, 11:37 PM   #14
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Quote:
Originally posted by Down With The King:
x^3+2x^2+6x+3; 2i
2i isn't a root of this equation.

x^3+2x^2+6x+3; 2i

(2i)³ = 2³i³ = 8(-1)i = -8i
2(2i)² = 2(2²)i² = 8(-1) = -8
6(2i) = 12i
3

-8i - 8 + 12i + 3 = -5 + 4i != 0.


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Old 02-13-2005, 11:41 PM   #15
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This topic would be great for me and my physics homework.
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Old 02-14-2005, 01:07 AM   #16
 
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Quote:
Find the other roots of this equation
x^3+2x^2+6x+3; 2i
Um, I see no equation here.

And remember, "I'm-a Luigi, number one!"
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Old 02-14-2005, 07:55 AM   #17
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^ If you presume it set equal to zero, it is.


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Old 02-14-2005, 01:10 PM   #18
 
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Well, I could presume it was set equal to -3 or to a google and it would be an equation. If anything should be presumed, it would be to set it equal to some y [i.e. f(x)].

Pardon my pickiness, but sloppy math questions bug me. I don't mind sloppy work, but the question should at least be clear.

Oh, and Andre, it's more like 95% of math past Geometry, not just Alg 1. [img]smile.gif[/img]

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Old 02-14-2005, 03:40 PM   #19
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I SAID ANDI SOLVE.

Not you guys, I just threw in together two types of things we've been doing in Algebra, it was not meant to be this much discussed.
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Old 02-14-2005, 04:04 PM   #20
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^Too bad. I can't solve it, so I'm leaving it up to them.

And they will battle to the death until I say otherwise.
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