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Thread: 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:

    $\displaystyle b^2 = x^2 + x + 7$

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

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

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

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

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

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

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