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Math Help - Im stuck!

  1. #1
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    Im stuck!

    'm stuck! I'm trying to find x using the cosine rule from:



    The answer is; x = 7.23 from:



    How?

    In another question i have to find the value of x which has the minimum value:

    b^2 = x^2 + x + 7

    I'm guessing i have to complete the square? but b = b^2 so do i take the route of the whole thing?
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  2. #2
    MHF Contributor
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    For the first question, use the quadratic formula. The solutions for ax^2 + bx + c = 0 are x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}.

    For the second question, completing the square is a good idea. You will get b^2 = \left(x + \frac{1}{2}\right)^2 + \frac{27}{4}.

    Then the minimum value for b is \sqrt{\frac{27}{4}}.
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  3. #3
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    i don't understand how to get the x^2 - 4(\sqrt2-1)x - (29+8\sqrt2)

    And the second one gives the answer to be {\frac {1}{2}}?
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  4. #4
    MHF Contributor
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    Quote Originally Posted by Flexible View Post
    i don't understand how to get the x^2 - 4(\sqrt2-1)x - (29+8\sqrt2)

    And the second one gives the answer to be {\frac {1}{2}}?
    The solutions for x^2 - 4(\sqrt2-1)x - (29+8\sqrt2) = 0 are:

    x = \frac{4(\sqrt2 - 1) \pm \sqrt{(4(\sqrt2 - 1))^2 + 4(1)(29+8\sqrt2)}}{2}.

    For the second problem, the answer is x = -\frac{1}{2}.
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  5. #5
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    How do you get -\frac{1}{2} I.e  \sqrt\frac{27}{4} = -\frac{1}{2} How?
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