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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.



Vectorspace
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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!



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04 Jul 2013, 11:38 am

Salome wrote:
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]


You left out x = 0



Salome
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04 Jul 2013, 11:54 am

ruveyn wrote:
Salome wrote:
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]


You left out x = 0


Is that it?!?!
I've solved it again doing the parentheses first and now I've got
x1=0
x2=5
x3=-9
But that only works for -4
I'm still doing something wrong!



Vectorspace
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04 Jul 2013, 12:01 pm

Salome wrote:
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?

Possibly. But Wolfram Alpha is probably more user-friendly for plotting.

Salome wrote:
Is that it?!?!
I've solved it again doing the parentheses first and now I've got
x1=0
x2=5
x3=-9
But that only works for -4
I'm still doing something wrong!

Seems like you inserted another error ahead of the square root.



Salome
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04 Jul 2013, 1:54 pm

Vectorspace wrote:
Salome wrote:
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?

Possibly. But Wolfram Alpha is probably more user-friendly for plotting.

Salome wrote:
Is that it?!?!
I've solved it again doing the parentheses first and now I've got
x1=0
x2=5
x3=-9
But that only works for -4
I'm still doing something wrong!

Seems like you inserted another error ahead of the square root.


Bloody hell, Wolfram Alpha is fantastic!! !! !! 8O
I can see everything! It makes so much more sense to me now!



Salome
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04 Jul 2013, 2:53 pm

Yes cracked it! Woho!! !! :bounce:



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04 Jul 2013, 5:00 pm

Dhp wrote:
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.


It really doesn't matter if you divide the 4 by the 2 before squaring or not. It's always the same answer either way, as long as you square both the numerator and denominator. In this case (4/2)^2 = 16/4 = 4, which is the same as (4/2)^2 = 2^2 = 4. It's really the next line where she made the mistake. She seems to of multiplied the fraction by 2 instead of squaring it and then for some reason divided the 5 by 2. Otherwise she would of gotten the 9 underneath the square root.



Salome
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04 Jul 2013, 5:21 pm

Jono wrote:
Dhp wrote:
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.


It really doesn't matter if you divide the 4 by the 2 before squaring or not. It's always the same answer either way, as long as you square both the numerator and denominator. In this case (4/2)^2 = 16/4 = 4, which is the same as (4/2)^2 = 2^2 = 4. It's really the next line where she made the mistake. She seems to of multiplied the fraction by 2 instead of squaring it and then for some reason divided the 5 by 2. Otherwise she would of gotten the 9 underneath the square root.


Thank you for clarifying!



Salome
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04 Jul 2013, 7:39 pm

Please please please please tell me I did this one right!

[img][800:576]http://farm4.staticflickr.com/3768/9213120544_7a656a0f8f_b.jpg[/img]



Vectorspace
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04 Jul 2013, 7:45 pm

Salome wrote:
Please please please please tell me I did this one right!

[img][800:576]http://farm4.staticflickr.com/3768/9213120544_7a656a0f8f_b.jpg[/img]

That's right (aside from the typo in line 5).

I'm just a little surprised that your assignment covers complex numbers, because they're usually taught way later than quadratic equations.



Salome
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04 Jul 2013, 8:01 pm

Vectorspace wrote:
Salome wrote:
Please please please please tell me I did this one right!

[img][800:576]http://farm4.staticflickr.com/3768/9213120544_7a656a0f8f_b.jpg[/img]

That's right (aside from the typo in line 5).

I'm just a little surprised that your assignment covers complex numbers, because they're usually taught way later than quadratic equations.


Yay!! ! Weeee! I'm not hopeless :D The typo, do you mean the line in the square root not covering the entire problem? I tried fixing that but it just wouldn't.
Well we don't cover them we have just been told that they exist. I'm not sure really, I'm studying long distance and don't have any classes. We are given assignments to turn in and to help us do them a couple of pages of lecture notes. And of course we have a book. To be honest though I've just looked in the book three times because it doesn't explain anything it just contains problems. The lecture notes don't really tell me anything either they sort of assume that you get this stuff.