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Salome
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02 Jul 2013, 7:54 am

ruveyn wrote:
Salome wrote:
Ok I'm done feeling sorry for myself for being completely hopeless and I've done some thinking.
I decided that it doesn't matter that I suck at maths and maybe even fail the course, I'm going to take it anyway. I mean I am learning new things and that was the whole point of taking the course in the first place.
Hopefully I haven't scared of all you math geniuses incase I need to ask more questions! :D


Ask. Just do not ask for the answers to your homework problems.

ruveyn


I wasn't aware that I did. I needed help breaking it down and every step explained (why it's done that way) I apologize if that was wrong!



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02 Jul 2013, 11:54 am

Aww...you haven't scared me at all. I understand how you feel. Remember, to be a success in life, you need to find out what you love to do, and find a way to make that a career. You're right; not all can be math geniuses (like VectorSpace), but if you find out what you love to do, and make that a career, you will find happiness and be able to spread that to others. I hope you do well in whatever you do.



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02 Jul 2013, 12:06 pm

Dhp wrote:
Aww...you haven't scared me at all. I understand how you feel. Remember, to be a success in life, you need to find out what you love to do, and find a way to make that a career. You're right; not all can be math geniuses (like VectorSpace), but if you find out what you love to do, and make that a career, you will find happiness and be able to spread that to others. I hope you do well in whatever you do.


Thank you! That's very sweet of you!
I have spent many years trying to figure out what my greatest love/interests are. I still don't know but I love technology, computers, art and writing. Anyhow I think mechanical engineering would be great! I would get to invent stuff and also design stuff.
However I would need to be great at maths to become an engineer which is not likely to ever happen but I figured I'd give it a shot.



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02 Jul 2013, 12:22 pm

I don't think you did anything bad. OK, you were given the solution, but you wanted to understand it.



Salome
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02 Jul 2013, 12:55 pm

Vectorspace wrote:
I don't think you did anything bad. OK, you were given the solution, but you wanted to understand it.


Well I have no interest in just getting the answer, I need and want to learn this stuff! I'm still hoping that my total cluelessness is just that, ignorant. There is still this glimmer of hope somewhere at the back of my head that thinks I might just be good at this stuff if I just learned it.



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03 Jul 2013, 2:44 pm

Ok I'm working on a third grade equation now. I know this is probably a long shot but, could I not just make it a second or ordinary equation and solve it like that?



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03 Jul 2013, 4:56 pm

Salome wrote:
Ok I'm working on a third grade equation now. I know this is probably a long shot but, could I not just make it a second or ordinary equation and solve it like that?

That sometimes works. If you already know one zero, you can do polynomial division and obtain a 2nd degree polynomial. But the approach to chose really depends on the equation.



Salome
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03 Jul 2013, 5:09 pm

Vectorspace wrote:
Salome wrote:
Ok I'm working on a third grade equation now. I know this is probably a long shot but, could I not just make it a second or ordinary equation and solve it like that?

That sometimes works. If you already know one zero, you can do polynomial division and obtain a 2nd degree polynomial. But the approach to chose really depends on the equation.


Ok so if I already have a one zero and make it into a 2nd degree then I will still end up with three roots right? Do I always need to account for all the roots and does it matter in which order?



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03 Jul 2013, 5:36 pm

Salome wrote:
Vectorspace wrote:
Salome wrote:
Ok I'm working on a third grade equation now. I know this is probably a long shot but, could I not just make it a second or ordinary equation and solve it like that?

That sometimes works. If you already know one zero, you can do polynomial division and obtain a 2nd degree polynomial. But the approach to chose really depends on the equation.


Ok so if I already have a one zero and make it into a 2nd degree then I will still end up with three roots right? Do I always need to account for all the roots and does it matter in which order?

In the "good" case, a 3rd degree polynomial has 3 real roots. But I think you should just post the problem here.



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03 Jul 2013, 5:42 pm

Salome wrote:
Ok so if I already have a one zero and make it into a 2nd degree then I will still end up with three roots right? Do I always need to account for all the roots and does it matter in which order?

Order doesn't matter.

You'll always get 3 roots with a 3rd degree equation, but sometimes you get a multiple root, and sometimes the roots are complex numbers. (An example of a multiple root: (x-1)(x-1) = x^2 - 2x + 1 = 0. x = 1 is the only solution, but the factor (x-1) shows up twice when you factor it.)


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Salome
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03 Jul 2013, 5:53 pm

Ok I get it. Thank you!



Salome
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03 Jul 2013, 6:24 pm

This feels wrong to me but I have no idea what I might have done wrong.

[img][800:576]http://farm4.staticflickr.com/3821/9205221636_c820a37c38_b.jpg[/img]



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03 Jul 2013, 8:43 pm

Hi, Salome. I hope you're doing well. I'm afraid that you have made three errors in this problem.

The first is that instead of dividing by x (even though all three terms had an x in common), you need to factor out the x so that you have this:

Note: for notational purposes, I will use the ^ sign to represent an exponent...like x^2 means x squared, and x^3 means x cubed.

The original problem then would look like this:

x^3 - 4x^2 - 5x = 0 (confusing, isn't it? Well, thanks to Bill Gates for not providing us with a good math text editor that can work anywhere instead of just Microsoft Word...heh)

Now, since all three terms have an x in common, we can factor out an x so it would now look like this:

x(x^2 - 4x - 5) = 0. Why the &$% would you do that? Because x = 0 is one solution to that equation. Look -> 0(0^2 - 4(0) - 5) = 0.

When you divided that whole thing by x, you eliminated one of the solutions. But please don't leave x= 0 in the cold cold snow! He has a wife and family! Why, everyday he has to walk 10 miles to the bus barefoot in the snow! Just kidding around...

The second mistake is that on the third line down on your step by step written out page, you have to follow the ordering of stuff. You need to divide 4 by 2 first before you square it under the square root. Then you'll get 2 squared, which is 4.

But wait, there's more (shuts off the TV because of the salesman's annoying voice)...

Underneath that radical, you will then have 4 + 5. But that equals 9, and the square root of 9 is 3. So now you should have this:

x = 0 (don't forget him), x = 2 +/- (that's plus or minus) 3. This means that your answer is 0, -1, and 5.

What? Is it possible to have three solutions? Of course it is. It is a cubic equation after all, and the most solutions you can have is 3. (jokingly) 3 is the number that you shalt count, and the number that you shalt count is 3...

I only hope that this helps you, Salome. If not, please let me know how I can help. Have a great evening.



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04 Jul 2013, 9:28 am

By the way, if you just want to check if your solution is right or if you want to know what the function "looks like", Wolfram Alpha is really helpful:
[url]http://www.wolframalpha.com/input/?i=x^3-4x^2-5x[/url]

I made the mistake to use a similar program for my homework in high-school because I was bored by the assignments, but it resulted in lots of arithmetical errors in the exams.



Salome
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04 Jul 2013, 9:32 am

Vectorspace wrote:
By the way, if you just want to check if your solution is right or if you want to know what the function "looks like", Wolfram Alpha is really helpful:
[url]http://www.wolframalpha.com/input/?i=x^3-4x^2-5x[/url]

I made the mistake to use a similar program for my homework in high-school because I was bored by the assignments, but it resulted in lots of arithmetical errors in the exams.


I haven't used a calculator for any of my assignments. I tried earlier today but I need to learn how it works. By the way, I have an old TI- 80 do you think it will do?



Salome
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04 Jul 2013, 10:28 am

Dhp wrote:
Hi, Salome. I hope you're doing well. I'm afraid that you have made three errors in this problem.

The first is that instead of dividing by x (even though all three terms had an x in common), you need to factor out the x so that you have this:

Note: for notational purposes, I will use the ^ sign to represent an exponent...like x^2 means x squared, and x^3 means x cubed.

The original problem then would look like this:

x^3 - 4x^2 - 5x = 0 (confusing, isn't it? Well, thanks to Bill Gates for not providing us with a good math text editor that can work anywhere instead of just Microsoft Word...heh)

Now, since all three terms have an x in common, we can factor out an x so it would now look like this:

x(x^2 - 4x - 5) = 0. Why the &$% would you do that? Because x = 0 is one solution to that equation. Look -> 0(0^2 - 4(0) - 5) = 0.

When you divided that whole thing by x, you eliminated one of the solutions. But please don't leave x= 0 in the cold cold snow! He has a wife and family! Why, everyday he has to walk 10 miles to the bus barefoot in the snow! Just kidding around...

The second mistake is that on the third line down on your step by step written out page, you have to follow the ordering of stuff. You need to divide 4 by 2 first before you square it under the square root. Then you'll get 2 squared, which is 4.

But wait, there's more (shuts off the TV because of the salesman's annoying voice)...

Underneath that radical, you will then have 4 + 5. But that equals 9, and the square root of 9 is 3. So now you should have this:

x = 0 (don't forget him), x = 2 +/- (that's plus or minus) 3. This means that your answer is 0, -1, and 5.

What? Is it possible to have three solutions? Of course it is. It is a cubic equation after all, and the most solutions you can have is 3. (jokingly) 3 is the number that you shalt count, and the number that you shalt count is 3...

I only hope that this helps you, Salome. If not, please let me know how I can help. Have a great evening.


Hello! Thank you for your explanation! I almost get it :D
However, could you write it out step by step like I did because I don't understand how you write out what you said. For me to understand this it needs to look the same as what I put up there but the correct way. I'm not good enough at maths to understand if I can't clearly see the connections.
Also now I'm thinking that one of my previous equations was solved wrong because I realised I didn't do the parentheses first grrrr!