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Math Help - Solutions of quadratic equation

  1. #1
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    Solutions of quadratic equation

    Hi having a bit of a mind blank trying to find the solutions to this quadratic equation, any hints would be invaluable

    x^2 -(5/3)x +4/3=0
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    Quote Originally Posted by offahengaway and chips View Post
    Hi having a bit of a mind blank trying to find the solutions to this quadratic equation, any hints would be invaluable

    x^2 -(5/3)x +4/3=0
    Have you tried using the formula ?
    Note that there are no real solutions to this equation.
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  3. #3
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    Yes, I keep getting math error on my calculator in the root part of the formula.
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    Quote Originally Posted by offahengaway and chips View Post
    Yes, I keep getting math error on my calculator in the root part of the formula.
    Run it complex mode.. The solution is complex.
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    Are there any fixed points on the graph of this equation, because that is what i am trying to answer
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    Quote Originally Posted by offahengaway and chips View Post
    Are there any fixed points on the graph of this equation, because that is what i am trying to answer
    The question is not clear.. What does fixed point of an equation mean?Fixed points are defined for functions (as far as I know).
    Also usually when one says "solve an equation", one should specify the domain... Is x real?

    So assuming we want x real, is the question asking for the fixed point of the function f(x) = x^2 -(5/3)x +4/3 ?
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    Yes, sorry, it is the fixed points of f(x)
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    Quote Originally Posted by offahengaway and chips View Post
    Yes, sorry, it is the fixed points of f(x)
    Then that means you have to solve f(x) = x.

    This means you have to actually solve x^2 -(8/3)x +4/3=0. This equation can be safely graphed to get real solutions
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  9. #9
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    x^2 -(5/3)x +4/3=0 is the equation from the function, unfortunately I have to find the answers algebraicly, that is where i am struggling.
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    Quote Originally Posted by offahengaway and chips View Post
    x^2 -(5/3)x +4/3=0 is the equation from the function, unfortunately I have to find the answers algebraicly, that is where i am struggling.
    First of all, do you understand how I got x^2 -(8/3)x +4/3=0 and secondly, do you know the quadratic formula?
    I am not understanding where you are getting stuck..
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  11. #11
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    I do know the quadratic formula, not too sure where you got the 8/3 from
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  12. #12
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    Quote Originally Posted by offahengaway and chips View Post
    I do know the quadratic formula, not too sure where you got the 8/3 from
    You are asked to find fixed points of f(x) = x^2 -(5/3)x +4/3. Read this, to know the definition.

    Now we know that a solution to f(x) = x is called a fixed point of f(x).
    So find values of x for which f(x) = x.

    f(x) = x implies x^2 -(5/3)x +4/3 = x, which implies x^2 -(8/3)x +4/3=0

    Now hit it with the quadratic formula and get x. Do you get it now?
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  13. #13
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    Sorry, I just cant see how the 8 comes into it
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  14. #14
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    Quote Originally Posted by offahengaway and chips View Post
    Sorry, I just cant see how the 8 comes into it


    Just basic algebraic manipulations:

    x^2 - \frac53 x + \frac43 = x \implies \left(x^2 - \frac53 x + \frac43 \right) - x = 0 \implies  x^2 - \left(\frac53 + 1 \right) x + \frac43 = 0 \implies x^2 - \frac83 x + \frac43 = 0<br />

    The only thing I have not explained is 1 + \frac53 = \frac83.
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  15. #15
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    AHAA, the penny has dropped thank you very much!!!!!!

    so simple, such a daft mistake
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