
Re:Help with Algebra??
It did not work. I will try again.
3=16x^2

Re:Help with Algebra??
the answer is
x = 0.5, try again.

Re:Help with Algebra??
It's not working. I either get 3=16x or 3=16x^2

Re:Help with Algebra??
First the bonus points:
1 2
 +  = 2
2x 4x
notice that this is the same as the above
1 1
 +  = 2
2x 2x
then:
2
 = 2
2x
4x = 2
x = 0.5
that was for the bonus points:
Here goes the universial method:
4x 4x
 +  = 2
8x² 8x²
8x
 = 2
8x²
16x² = 8x
16x = 8
x = 0.5

Easy!
I'm begining to think you are jerking my chain.

Re:Help with Algebra??
I don't know how to do that. That was not easy it is very hard. I did not understand anything on your last post. I have to tell you something that i never told you before. I'm a slow learner and i have a learning disability. How did i make it through algebra 1, 2 years ago it beats me.

Re:Help with Algebra??
Get tutoring, that's all I can say  it's MUCH easier to understand if you get tutoring in person.

Re:Help with Algebra??
Yo hold up. I got something better. Do you use yahoo messenger?? There is something called doodle it's and IMVironment where you could draw out your problems and answers. I use yahoo. I know i'm going to need a tutor when it comes down to Analytic Geometry & Calculus 1, 2 and 3. Also Linear algebra. Please reply back. Thanks.

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Re:Help with Algebra??
[quote author=Lovechild link=board=14;threadid=7355;start=30#msg67884 date=1058025366]
"crossed over"
1 * 2x 4*(x1)
 +  = 10
(x 1)*2x 2x*(x1)
Now we do the math  nothing changes but appearences
2x 4x  1
 +  = 10
2x²  2x 2x²  2x
[/quote]
Not to piss anyone off or anything, but this action should give a result of:
2x 4x  4
 +  = 10
2x²  2x 2x²  2x
So I guess the result in that one, isn't what you ended up with.
edit
And in the end you'll end up with solving the roots of a second degree equation, like 0=ax²+bx+c
Which either can have only one solution, two solutions or no solution at all.
The way to find these is by using the formular:
b + sqrt(b²4ac)
x = 
2a
If (b²4ac) gives a negative value, it has no solution, atleast not within the number group R, but if we use Z, then it has.

Re:Help with Algebra??
WTF... did I miscalculated also... oh I'm so embarrased now.
oh it's a minor typo I see, I should have catched that as well.
I think the Z domain is a tad out of scope for the project at hand, but true it's an awesome tool.
But you are of course correct about that root solving thing, I'm guessing I'm chalk this up to picking a bad example, it was maybe a tad to advanced (even for me it seems  I glad stumbled into the common trap, forgetting prior knowledge)
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