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Thread: cosine rule

  1. #1
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    cosine rule

    hello,
    i have this

    in triangle ABC, AB=(5-x) BC=(4+x) , angle ABC = 120 degrees and AC = y

    i know that y^2=x^2 - x + 61

    now i have to complete the square to find minimum value of y^2 and give value of x for which this occurs


    but i cant figure out how to start with it, because there is no roots
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  2. #2
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    Re: cosine rule

    You start by doing as instructed, complete the square...
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  3. #3
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    Re: cosine rule

    Minimum value for $y^2$ occurs at $x=\dfrac{-b}{2a}=\dfrac{1}{2}$ ...
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  4. #4
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    Re: cosine rule

    I have no idea what "no roots" has to do with "completing the square". A "perfect square" is of the form (x- a)^2= x^2- 2ax+ a^2. Here, you have x^2- x+ 61. The coefficient of x is 2a= -1 so a= -1/2. Then a^2= \frac{1}{4}. To get a "perfect square" we must add (and subtract) -1/4:
    [tex]x^2- x+ 61= x^2- x+ \frac{1}{4}- \frac{1}{4}+ 61= (x- \frac{1}{2})^2+ \frac{243}{244}[tex].

    y^2= (x- \frac{1}{2})^2+ \frac{243}{244}.

    What is the smallest value of the right side and for what value of x does that occur.
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  5. #5
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    Re: cosine rule

    thanks , seems i misunderstood the question somehow.. smallest value is 60.75
    but what is that [tex]\frac{243}{244}[tex] ?
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  6. #6
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    Re: cosine rule

    must be a typo ... and the smallest positive value for $y^2$ is 60.75

    $y^2 = x^2 - x + 61$

    $y^2 = \left(x^2 - x + \dfrac{1}{4}\right) + 61 - \dfrac{1}{4}$

    $y^2 = \left(x - \dfrac{1}{2}\right)^2 + 61 - \dfrac{1}{4}$

    $y^2 = \left(x - \dfrac{1}{2}\right)^2 + \dfrac{244}{4} - \dfrac{1}{4}$

    $y^2 = \left(x - \dfrac{1}{2}\right)^2 + \dfrac{243}{4}$

    $y_{min} = \dfrac{9\sqrt{3}}{2}$
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