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    (Original post by KittyRe-play)
    But then using the y method the solutions are x=3 & x=2 and it's not the same as 4 and 9. So is that incorrect?
    no, using the y-method you got y=3 and y=2
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    (Original post by TenOfThem)
    no, using the y-method you got y=3 and y=2
    Yeah so why do you square them up to get 4 and 9? Is it because y^2=x? What is this general formula I've never come across it before?
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    (Original post by KittyRe-play)
    Yeah so why do you square them up to get 4 and 9? Is it because y^2=x? What is this general formula I've never come across it before?
    I do not know what formula you are referring to

    The equation was

    x - 5\sqrt{x} + 6 = 0

    You chose to solve

    y^2 - 5y + 6 = 0 instead


    Can you not see how they are linked
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    (Original post by TenOfThem)
    I do not know what formula you are referring to

    The equation was

    x - 5\sqrt{x} + 6 = 0

    You chose to solve

    y^2 - 5y + 6 = 0 instead


    Can you not see how they are linked
    No I can see how they are linked. Basically you have replaced x with y squared. But how can you just plug y squared in out of nowhere is what I'm asking, like it must be a formula or something right?
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    the equation, when everything is put on the LHS, is what is termed a "disguised quadratic" - as it`s just that - a quadratic, in some variable, in disguise.

    you can either, choose method 1). re-arranging and squaring, which may or may not introduce entraneous solutions,

    or method 2). as TenOfThem has explained, use something like p=\sqrt{x}, p^{2}=x, solve for p, and then square the result, then check in the original equation.
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    (Original post by Hasufel)
    the equation, when everything is put on the LHS, is what is termed a "disguised quadratic" - as it`s just that - a quadratic, in some variable, in disguise.

    you can either, choose method 1). re-arranging and squaring, which may or may not introduce entraneous solutions,

    or method 2). as TenOfThem has explained, use something like p=\sqrt{x}, p^{2}=x, solve for p, and then square the result, then check in the original equation.
    So is p=\sqrt{x}, p^{2}=x some sort of formula? Like a general formula? What is it specifically used for?
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    (Original post by KittyRe-play)
    So is p=\sqrt{x}, p^{2}=x some sort of formula? Like a general formula? What is it specifically used for?
    Nah, just a general substitution which is "convenient" for transforming this particular equation into a quadratic.

    e.g. say you had:

    e^{2x}-e^{x}-2=0

    you should, at that stage, be practiced enough to recognise that there`s a convenient substitution:

    e^{x}=p (or r, s t, or any other "arbitrary" letter you choose to solve the quadratic for)

    so you get:

    p^{2}-p-2=0

    solve for p, then as the substitution demands, take natural logs to solve for x,,,
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    (Original post by Hasufel)
    Nah, just a general substitution which is "convenient" for transforming this particular equation into a quadratic.

    e.g. say you had:

    e^{2x}-e^{x}-2=0

    you should, at that stage, be practiced enough to recognise that there`s a convenient substitution:

    e^{x}=p (or r, s t, or any other "arbitrary" letter you choose to solve the quadratic for)

    so you get:

    p^{2}-p-2=0

    solve for p, then as the substitution demands, take natural logs to solve for x,,,
    Thanks man that makes so much sense now.
 
 
 
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