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Math Help - How to turn this periodic function in a quadratic equation?

  1. #1
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    How to turn this periodic function in a quadratic equation?

    I have acquired the equation:

    0 = 70tanθ - 1 / 15.3125 (1+tan θ)

    I understand that I can solve this by turning it into a quadratic, just unsure how you rearrange this so that it is a quadratic.

    Any help would be extremely appreciated (:
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  2. #2
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    Quote Originally Posted by eriiin View Post
    I have acquired the equation:

    0 = 70tanθ - 1 / 15.3125 (1+tan θ)

    I understand that I can solve this by turning it into a quadratic, just unsure how you rearrange this so that it is a quadratic.

    Any help would be extremely appreciated (:
    I can't read this. Please either use LaTeX or brackets where they're needed.

    Is it 0 = 70\tan{\theta} - \frac{1}{15.3125(1 + \tan^2{\theta})}?
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  3. #3
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    Yes sorry, that is correct.
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  4. #4
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    Start by getting a common denominator:

    0 = 70\tan{\theta} - \frac{1}{15.3125(1 + \tan^2{\theta})}

    0 = \frac{1071.875\tan{\theta}(1 + \tan^2{\theta})}{15.3125(1 + \tan^2{\theta})} - \frac{1}{15.3125(1 + \tan^2{\theta})}

    0 = \frac{1071.875\tan{\theta}(1 + \tan^2{\theta}) - 1}{15.3125(1 + \tan^2{\theta})}

    0 = 1071.875\tan{\theta}(1 + \tan^2{\theta}) - 1

    0 = 1071.875\tan^3{\theta} + 1071.875\tan{\theta} - 1.


    You may need technology to solve this equation.
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  5. #5
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    I am unsure to whether I am doing this correctly, because I've never done a quadratic that has a number with a cube root before.

    So I am guessing I need to somehow get the:

    .

    to a tan^2.

    Any guidance to how I do this?
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  6. #6
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    Perhaps you could solve it as a "cubic" instead of a "quadratic"

    here's the cubic formula:

    Cubic function - Wikipedia, the free encyclopedia

    hope that helps
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  7. #7
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    Never even seen that equation before /:
    But I only have three numbers which I can use, not four, so I take it that idea isn't possible.
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  8. #8
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    Quote Originally Posted by eriiin View Post
    Never even seen that equation before /:
    But I only have three numbers which I can use, not four, so I take it that idea isn't possible.
    If you put:

    a = 1071.875 ( tan^3)

    b = 0 ( tan^2)

    c = 1071.875 ( tan)

    d = -1

    so it can be solved the " tan^2" doesn't appear but it exists (this is true in all polynomials)

    I hope that explains it
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