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Thread: Solve for W

  1. #1
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    Solve for W

    Solve for W

    $\displaystyle
    V=\sqrt{2Lw+w^2}
    $
    My answer was $\displaystyle
    \Longrightarrow
    w=V^2-2L
    $
    But apparently thats wrong according to my teacher.
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  2. #2
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    Quote Originally Posted by OzzMan View Post
    Solve for W

    $\displaystyle
    V=\sqrt{2Lw+w^2}
    $
    My answer was $\displaystyle
    \Longrightarrow
    w=V^2-2L
    $
    But apparently thats wrong according to my teacher.
    Square both sides of the equation:

    $\displaystyle V^2=2Lw + w^2~\iff~w^2 + 2Lw - V^2 = 0$

    This is a quadratic equation. Use the formula to solve it.
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  3. #3
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    What do you mean exactly? What I did was square both sides like you said and got $\displaystyle V^2=2Lw+w^2 \Longrightarrow w^2=V^2 - 2Lw $
    Then I did $\displaystyle \frac{w^2}{w}=V^2 - \frac{2Lw}{w} $
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  4. #4
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    Quote Originally Posted by OzzMan View Post
    What do you mean exactly? What I did was square both sides like you said and got $\displaystyle V^2=2Lw+w^2 \Longrightarrow w^2=V^2 - 2Lw $
    Then I did $\displaystyle \frac{w^2}{w}=V^2 - \frac{2Lw}{w} $
    Unfortunately you forgot to divide the Vē by w.

    If you have a quadratic equation :

    $\displaystyle ax^2+bx+c = 0$

    Then this equation has the solutions:

    $\displaystyle w = \frac{-b \pm \sqrt{b^2-4ac}}{2a}$

    With your problem you have:

    a = 1
    b = -2L
    c = -Vē

    Plug in these terms into the formula to get the solutions.

    (For confirmation only: $\displaystyle w = L\pm\sqrt{L^2+V^2}$
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  5. #5
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    Just making sure I did this right
    $\displaystyle w=\frac{2L\pm\sqrt{4L^2+4v^2}}{2}$
    How did you get $\displaystyle w=L \pm\sqrt{L^2+V^2} $
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  6. #6
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    $\displaystyle \sqrt {4L^2 + 4w^2 } = \sqrt {4\left( {L^2 + w^2 } \right)} = \sqrt 4 \sqrt {\left( {L^2 + w^2 } \right)}
    $
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  7. #7
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    shouldn't it be a $\displaystyle -2L$ instead of $\displaystyle 2L$? According to the equation $\displaystyle w^2+2Lw-V^2=0$
    $\displaystyle
    w=\frac{-2L\pm\sqrt{4L^2+4v^2}}{2}
    $
    So the answer would be $\displaystyle
    w = -L\pm\sqrt{L^2+V^2}
    $
    Last edited by OzzMan; Feb 8th 2008 at 01:42 PM.
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  8. #8
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    Quote Originally Posted by OzzMan View Post
    shouldn't it be a -2L instead of 2L? According to the equation $\displaystyle w^2+2Lw-V^2=0$
    $\displaystyle
    w=\frac{-2L\pm\sqrt{4L^2+4v^2}}{2}
    $
    So the answer would be $\displaystyle
    w = -L\pm\sqrt{L^2+V^2}
    $
    Yes. Good pick-up. Earboth made a small sign error when he substituted for b:

    Quote Originally Posted by earboth View Post
    [snip]
    With your problem you have:

    a = 1
    b = -2L
    c = -Vē

    Plug in these terms into the formula to get the solutions.

    (For confirmation only: $\displaystyle w =$ L$\displaystyle \pm\sqrt{L^2+V^2}$
    Last edited by mr fantastic; Feb 8th 2008 at 04:21 PM. Reason: Fixed the color
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