Anyone good with fractions? - Page 2

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Anyone good with fractions?

This is a discussion on Anyone good with fractions? within the General Off Topic Discussion forums, part of the Life Beyond Horses category

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        09-01-2008, 05:46 AM
    Good maths teachers are so important. People feel ashamed if they have difficulty with reading and writing, but it's somehow 'okay' to be bad at maths and science because they're 'hard'. Sure, not everyone has a natural aptitude, but they are no harder than achieving a good level of literacy. And teachers have a lot to do with that attitude, unfortunately.

    If you've got any questions, let us know! I try and find several ways of explaining things ... for me, the proof that I understand something is if I can teach it, and I have my own way of helping myself understand something. I'll put it into my own language, so to speak - describing stuff with 'thingy' and 'this bit has to do that so the other bit can eat it up' - and then put it into the correct terms. I'm also really picky about being precise with your language, because in science and maths each word has such a distinct meaning that if you waffle you'll only confuse yourself and others.
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        09-01-2008, 01:14 PM
    I am one of those where writing comes easy, speaking does not, and I feel like a complete and utter failure when I can't comprehend a math problem. I just want to cry.

    I think my new strategy will be to not write the long written notes, and just make more detailed example notes.

    I will indeed come for more help. Thanks!
        09-01-2008, 01:37 PM
    Examples make everything better :)
        09-10-2008, 07:53 PM
    Okay my masters, new issue.

    Solve by "Completing the Square". Write your answers in both exact form and approximate form, rounded to the hundredths place. If there are no solutions, so state.

    My problem:

    M^2 + 3m = 1

    I have the answers photocopied from the answer textbook. What I am trying to do is work them out before I cheat and check, and have done fairly well on other methods. My notes have turned to gibberish when I try to read about this though.

    Do you want the work so you can explain it?
        09-10-2008, 08:08 PM
    Completing the square ... haven't done that in about nine or ten months so this will be rough xD

    m^2 + 3m - 1 = 0
    (I always put it in this form - something = 0 - because that is how it's the most familiar for me. It's not actually necessary :) Now, just to keep our algebraic muscles working, let's consider this as am^2 + bm + c, where a = 1, b = 3 and c = -1, so when I refer to the a term, b term and c term, that's what I mean. You should be used to considering quadratic equations in that general form.)

    Divide the b-term by two and put it into the squared bit, and see what that comes out as:

    (m + 3/2)^2 = m^2 + 3m + (3/2)^2

    (3/2)^2 = (3^2)/(2^2) = 9/4

    Which means that:

    m^2 + 3m -1 = m^2 + 3m + 9/4 - 9/4 - 1

    If you look at that: 9/4 - 9/4 = 0, yup? So both sides of that equation have the same value. But we now also have the (m + 3/2)^2 bit hanging around in there, in italics! Therefore:

    m^2 + 3m - 1 = (m + 3/2)^2 - 9/4 - 1

    M^2 + 3m - 1 = (m + 3/2)^2 - 13/4


    (m + 3/2)^2 - 13/4 = 0

    (m + 3/2)^2 = 13/4

    M + 3/2 = +- SQRT (13/4)

    Never forget that +/-, it's very important!

    Once you understand the method, you quickly start to do something like this:

    X^2 + bx + c = 0

    Well, I know how that works, so I just automatically switch it to:

    (x + b/2)^2 = -c - b^2/4
    X + b/2 = +- SQRT (-c - b^2/4)
    And solve from there.

    Completing the square is how we get the quadratic formula, too :)


    Our final answer is that:

    m + 3/2 = +- SQRT (13/4)
    M1 = (SQRT13)/2 - 3/2
    M2 = -(SQRT13)/2 - 3/2

    Exact form would be as above: SQRT13 is a surd and hence irrational. Approximate form would be to stick that into your calculator and see what number it gives you.

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