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Salome
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30 Jun 2013, 3:37 pm

Vectorspace wrote:
Salome wrote:
But see thats the thing, I don't. How does 9/4 =5?
I'm sorry if I'm hopeless!

Well, you could just type the expression above in a calculator, but I'm glad that you're trying to understand it. So this is the long version:
Image


Aaaaaah ok I get it! I missed a few steps!
Another question, how do I know which method to perform? I know that you can tell what kind of equation it is by the exponent but then what?



Vectorspace
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30 Jun 2013, 3:49 pm

Salome wrote:
Aaaaaah ok I get it! I missed a few steps!
Another question, how do I know which method to perform? I know that you can tell what kind of equation it is by the exponent but then what?

If the exponent is 3 or greater, then, in the general case, you don't have a chance. The method that is usually taught at school is just guessing a zero (in the hope that it's an integer) and doing polynomial division. Sometimes you can also factor out or substitute something.

OK, there is Cardano's method, but that is rarely applied nowadays. Using a computer is so much easier.



Salome
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30 Jun 2013, 3:57 pm

Vectorspace wrote:
Salome wrote:
Aaaaaah ok I get it! I missed a few steps!
Another question, how do I know which method to perform? I know that you can tell what kind of equation it is by the exponent but then what?

If the exponent is 3 or greater, then, in the general case, you don't have a chance. The method that is usually taught at school is just guessing a zero (in the hope that it's an integer) and doing polynomial division. Sometimes you can also factor out or substitute something.

OK, there is Cardano's method, but that is rarely applied nowadays. Using a computer is so much easier.


Ah ok! i'll be learning about third grade equations soon I'm guessing that will be interesting :pale:



Salome
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30 Jun 2013, 4:40 pm

Is this correct?

[img][800:576]http://farm6.staticflickr.com/5346/9179054138_571bda43c7_b.jpg[/img]



Vectorspace
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30 Jun 2013, 6:18 pm

Salome wrote:
Is this correct?

[img][800:576]http://farm6.staticflickr.com/5346/9179054138_571bda43c7_b.jpg[/img]

First of all, the signs are wrong. If you calculate f(5) in the last line, you will get 30, not 0, and if you put x=–2, the result will be –12.
Secondly, as stated by Stargazer43 before, you're dropping the factor 2 by dividing. If you put it back, you get:

f(x)=2*(x–5)*(x+2)

When unsure, just expand the expression and check if it matches the original definition.



Salome
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30 Jun 2013, 6:24 pm

Ok so it's just the last line that is wrong?



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30 Jun 2013, 6:28 pm

Salome wrote:
Ok so it's just the last line that is wrong?

Yes.

In the 4th last line, you should have written x₁ rather than x, because you only write + and not ±.



Salome
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30 Jun 2013, 6:31 pm

Vectorspace wrote:
Salome wrote:
Ok so it's just the last line that is wrong?

Yes.

In the 4th last line, you should have written x₁ rather than x, because you only write + and not ±.


Ah ok I get it!
Thank you! :D



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30 Jun 2013, 6:43 pm

Okay, Salome - you're almost right except for one thing: the last line. When x = 5, and x = -2, what you need to do is this: subtract 5 from the first one to be x-5 = 0. For the second one, you need to subtract -2 from the sex...oops I mean second one (hee hee hee) to be x -(-2) = 0, and remember that negative times negative is positive. That means x+2 = 0. Why all of this? Remember that rule - if ab = 0 then a = 0 or b = 0, or both? Check this out.

x - 5 = 0, or if a = x - 5, then a = 0

x + 2 = 0, or if b = x + 2, then b = 0,

Now, we know that 0 times 0 is 0...we can multiply x - 5 by x + 2, and get...

(x - 5)(x + 2) = 0, and voila (confetti falls from the sky)! And we have our equation! Don't believe me? Multiply it back using FOIL and you will get x squared minus 3x minus 10 = 0, and since 2 times zero is zero, we could multiply both sides by 2 to get the original problem.


Note: I could have stated that equation as a function, for it is one. Yeah, we have two different x values, but each x value will produce a distinct y, even though it could be the same y value. You can have that situation occur in a function. (For vectorspace: and I know that this function will not have an inverse. :-) Thank you for correcting me, as I didn't realize that the square root is DEFINED to be a function.)



Salome
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30 Jun 2013, 7:07 pm

Dhp wrote:
Okay, Salome - you're almost right except for one thing: the last line. When x = 5, and x = -2, what you need to do is this: subtract 5 from the first one to be x-5 = 0. For the second one, you need to subtract -2 from the sex...oops I mean second one (hee hee hee) to be x -(-2) = 0, and remember that negative times negative is positive. That means x+2 = 0. Why all of this? Remember that rule - if ab = 0 then a = 0 or b = 0, or both? Check this out.

x - 5 = 0, or if a = x - 5, then a = 0

x + 2 = 0, or if b = x + 2, then b = 0,

Now, we know that 0 times 0 is 0...we can multiply x - 5 by x + 2, and get...

(x - 5)(x + 2) = 0, and voila (confetti falls from the sky)! And we have our equation! Don't believe me? Multiply it back using FOIL and you will get x squared minus 3x minus 10 = 0, and since 2 times zero is zero, we could multiply both sides by 2 to get the original problem.
Thank you for taking the time to answer!
However now I'm confused again.

Note: I could have stated that equation as a function, for it is one. Yeah, we have two different x values, but each x value will produce a distinct y, even though it could be the same y value. You can have that situation occur in a function. (For vectorspace: and I know that this function will not have an inverse. :-) Thank you for correcting me, as I didn't realize that the square root is DEFINED to be a function.)



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30 Jun 2013, 7:14 pm

Please tell me where you're confused, Salome. I am sorry to confuse you.



Salome
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30 Jun 2013, 8:04 pm

Dhp wrote:
Please tell me where you're confused, Salome. I am sorry to confuse you.


It's ok!
It's not your fault that I'm hopeless! In my own defense though, I dropped out of school somewhere around fifth grade so needless to say the gaps in my knowledge are huge.
I was cooking so I didn't see your answer before.
The last step, should I multiply by two?



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30 Jun 2013, 9:06 pm

Salome wrote:
Is this correct?

[img][800:576]http://farm6.staticflickr.com/5346/9179054138_571bda43c7_b.jpg[/img]


Are you going to buy that Math Type package?

ruveyn



Salome
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30 Jun 2013, 9:16 pm

ruveyn wrote:
Salome wrote:
Is this correct?

[img][800:576]http://farm6.staticflickr.com/5346/9179054138_571bda43c7_b.jpg[/img]


Are you going to buy that Math Type package?

ruveyn


:D no I can't afford it but I've been assured by my teachers that the free version works well for what we will be doing.
I'm feeling very discouraged though. I'm not sure I will continue with the course. There is too much that I don't know and my brain is very unreliable. Due to fatigue it's not online most days. When I applied for the course I had just started new medication and was doing really well but it only lasted for a couple of weeks and now I'm back to being dead tired all the time.



Salome
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02 Jul 2013, 1:37 am

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



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02 Jul 2013, 7:08 am

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