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Size:  336.8 KB need help with b) as I font get how you can obtain the equation shown inthr question . This is as far as I get Attachment 611596611598 thanks
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    (Original post by coconut64)
    Name:  14844155787691770128893.jpg
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Size:  336.8 KB need help with b) as I font get how you can obtain the equation shown inthr question . This is as far as I get s
    Subtract x/2 from both sides and then...
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    (Original post by Gregorius)
    Subtract x/2 from both sides and then...
    But that gives you this : 7/x^2 - x/2 = 0 but the question shows that there should be a x instead of 0 and there should be a positive sign ...

    Thanks
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    (Original post by coconut64)
    But that gives you this : 7/x^2 - x/2 = 0 but the question shows that there should be a x instead of 0 and there should be a positive sign ...

    Thanks
    You need to remember that you are solving for p. The fact that you have stuck 7 there shows that you have done it! Your equation is equivalent to the original equation, as required.
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    (Original post by Gregorius)
    You need to remember that you are solving for p. The fact that you have stuck 7 there shows that you have done it! Your equation is equivalent to the original equation, as required.
    That would be a guess though as I have - in my equation but in the question it is +

    Also I am confused with what you meant by solving for p , there is no p ?

    Thanks
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    (Original post by coconut64)
    That would be a guess though as I have - in my equation but in the question it is +

    Also I am confused with what you meant by solving for p , there is no p ?

    Thanks
    The way you can do this and this has always worked for me, is by working backwards and then swap it around, so  x = \frac{p}{x^2} +\frac{x}{2}

    Then I would want to multiply both sides by 2x^2 to get  2x^3 = 2p +x^3 then  x^3 = 2p so you can find what p is, then you can use that and form the equation, I hope I haven't shown too much here.
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    (Original post by NotNotBatman)
    The way you can do this and this has always worked for me, is by working backwards and then swap it around, so  x = \frac{p}{x^2} +\frac{x}{2}

    Then I would want to multiply both sides by 2x^2 to get  2x^3 = 2p +x^3 then  x^3 = 2p so you can find what p is, then you can use that and form the equation, I hope I haven't shown too much here.
    Great advice. Working backwards for these questions can be very useful.
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    (Original post by notnek)
    Great advice. Working backwards for these questions can be very useful.
    If I am to construct the equation, why can I do so? It is not working out for me; I am struggling to construct the same equation as the one shown in the question. How do I do it not considering the backward method? Thanks
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    (Original post by coconut64)
    If I am to construct the equation, why can I do so? It is not working out for me; I am struggling to construct the same equation as the one shown in the question. How do I do it not considering the backward method? Thanks
    NotNotBatman has shown you how to work backwards. Quote him if you don't understand part of his post.
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    (Original post by notnek)
    NotNotBatman has shown you how to work backwards. Quote him if you don't understand part of his post.
    I understand his method but I am just wondering if you know how to construct the equation without using the backward method. Cheers
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    (Original post by coconut64)
    I understand his method but I am just wondering if you know how to construct the equation without using the backward method. Cheers
    Do the backwards method backwards...
    that's how you would construct it, but it can be difficult to see.
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    (Original post by coconut64)
    I understand his method but I am just wondering if you know how to construct the equation without using the backward method. Cheers
    Sorry I misread you post.

    From the working in your first post, try adding \frac{x}{2} to both sides. It's possible that Gregorius meant this instead of subtracting from both sides.
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    (Original post by coconut64)
    I understand his method but I am just wondering if you know how to construct the equation without using the backward method. Cheers
    Divide both sides by x^2 then do \frac{14}{x^2}=\frac{p}{x^2} + \frac{x}{2} and solve for p. Flows nicely this way.
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    (Original post by coconut64)
    Also I am confused with what you meant by solving for p , there is no p ?
    The question says that you should show that <something> can be written in the form <something else involving p> where p is to be found. You are "solving for p" in the sense that you have to find p so that the identity is true.

    Others in this thread have suggested ways to approach this question, so I will just note that when you get a question phrased in this way, you have to read the question setter's "code" - questions written in this form are giving you the hint to start from the end (with the expression involving p) and work back to the original expression, finding a value of p that makes it work.

    But I also notice that you ask the excellent question:

    I understand his method but I am just wondering if you know how to construct the equation without using the backward method.
    The question setter had to construct the equation involving p without knowing in advance the form of that expression - how did he or she do it? The answer lies in the surrounding context of the question. You are being asked to construct iteration schemes that will converge to numerical solutions to equations. The equation setter has one great advantage over you - they know the conditions under which such iteration schemes actually converge to the right answer! Therefore they can look for equations of the right form that do as they want - there is a degree of trial and error in this sort of construction, but once you know roughly what you're looking for, things tend to drop out relatively easily.

    However, in answering the question, you don't necessarily know the direction that you need to set off in - hence you should use the "working backwards from the form of the answer" method (as suggested by the "code" of the question).
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    So it is okay to start using the the equation given in this question, even in exams as I don't really see this backward method in mark scheme normally. Thanks
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    (Original post by coconut64)
    So it is okay to start using the the equation given in this question, even in exams as I don't really see this backward method in mark scheme normally. Thanks
    Yes.
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    Thanks for the help everyone
 
 
 
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