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Math Help - Discriminants 2

  1. #1
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    Unhappy Discriminants 2


    Show that mx - (m + 1)x + 1 = 0 will have rational roots if m is an element of Q (I think Q means quotient?).

    Could someone please show me how to do this?
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  2. #2
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    Quote Originally Posted by xwrathbringerx View Post
    Show that mx - (m + 1)x + 1 = 0 will have rational roots if m is an element of Q (I think Q means quotient?).

    Could someone please show me how to do this?
    Have you calculated the discriminant? What do you notice? Therefore, what are the roots ....? By the way, Q is the symbol for rational numbers .....
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  3. #3
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    Can the solution be like this:

    For rational roots, discriminant = square no.

    Discriminant = (- m - 1)^2 - 4(m)(1)

    ... (working out) ...

    = (m -1)^2

    m -1 = (/q + 3)^2
    = (p+3q)^2 / q^2
    = rational square no.
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  4. #4
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    Quote Originally Posted by xwrathbringerx View Post
    Can the solution be like this:

    For rational roots, discriminant = square no.

    Discriminant = (- m - 1)^2 - 4(m)(1)

    ... (working out) ...

    = (m -1)^2

    m -1 = (/q + 3)^2
    = (p+3q)^2 / q^2
    = rational square no.
    x = \frac{m+1 \pm |m - 1|}{2} \in Q if m \in Q.
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  5. #5
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    Errr sori but what exactly do you mean? Which part of my solution am I meant to fix?
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  6. #6
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    Quote Originally Posted by xwrathbringerx View Post
    Discriminant = (- m - 1)^2 - 4(m)(1)

    ... (working out) ...

    = (m -1)^2

    m -1 = (/q + 3)^2
    = (p+3q)^2 / q^2
    = rational square no.
    Quote Originally Posted by xwrathbringerx View Post
    Errr sori but what exactly do you mean? Which part of my solution am I meant to fix?
    What I mean is that I have no idea where the stuff in red has come from or what it means.
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  7. #7
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    Oops I meant:

    Assume that m is rational.
    Therefore, m can be written in the form p/q where p and q are positive integers with no common factor.
    Hence, (m-1)^2 = (p/q -1)^2
    = (p-q)^2 / q^2
    = rational square no.
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  8. #8
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    Quote Originally Posted by xwrathbringerx View Post
    Oops I meant:

    Assume that m is rational.
    Therefore, m can be written in the form p/q where p and q are positive integers with no common factor.
    Hence, (m-1)^2 = (p/q -1)^2
    = (p-q)^2 / q^2
    = rational square no.
    But how is that relevant to answering the question?
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  9. #9
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    Don't I have to prove (m-1)^2 is rational?

    The only way I could think of doing that was to prove m was rational.
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  10. #10
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    Quote Originally Posted by xwrathbringerx View Post
    Don't I have to prove (m-1)^2 is rational? Mr F says: Where does the question ask this?

    The only way I could think of doing that was to prove m was rational.
    Here is a reminder of the question and a reply I gave:

    Quote Originally Posted by xwrathbringerx View Post
    Show that mx - (m + 1)x + 1 = 0 will have rational roots if m is an element of Q (I think Q means quotient?).

    [snip]
    Quote Originally Posted by mr fantastic View Post
    x = \frac{m+1 \pm |m - 1|}{2} \in Q if m \in Q.
    Think about what the question is asking and what my answer is saying.
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