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Thread: completing square

  1. #1
    Junior Member
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    completing square

    help........

    i have Three questions which i really don't know how to do
    pls help me!!
    solve the given quadratic equations by completing the square.
    $\displaystyle
    (1)----v(v + 2) = 15
    $
    ---------------------------------------------------------------------
    $\displaystyle
    (2)----5T^2 - 10T + 4 = 0
    $

    $\displaystyle (3)$In machine design. in finding the outside diameter$\displaystyle D_o$ of a hollow shaft, the equation$\displaystyle D_o^2 - DD_o - 0.25D^2 = 0$ is used Solve for $\displaystyle D_o$ if $\displaystyle D = 3.625cm$
    Last edited by danielwu; Sep 7th 2006 at 12:03 PM.
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  2. #2
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by danielwu
    $\displaystyle
    (1)----v(v + 2) = 15
    $
    $\displaystyle v^2 + 2v = 15$

    Now, $\displaystyle x^2 + 2ax + a^2 = (x + a)^2$

    Here we have 2 = 2a so a = 1. Thus add $\displaystyle a^2 = 1$ to both sides of the equation:
    $\displaystyle v^2 + 2v + 1 = 15 + 1$

    $\displaystyle (v + 1)^2 = 16$ Take the square root of both sides.

    $\displaystyle v + 1 = \pm 4$

    $\displaystyle v = -1 \pm 4 = $ 3 or -5.

    (You should always be in the habit of checking your solutions to make sure they work. As it happens, they both do.)

    -Dan
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  3. #3
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by danielwu
    (2)----5T^2 - 10T + 4 = 0
    [/tex]
    $\displaystyle 5T^2 - 10T = -4$

    $\displaystyle T^2 - 2T = -\frac{4}{5}$

    (See previous post)
    -2 = 2a, so a = -1. Thus add $\displaystyle a^2 = 1$ to both sides.

    $\displaystyle T^2 - 2T + 1 = -\frac{4}{5} + 1$

    $\displaystyle (T - 1)^2 = \frac{1}{5}$

    $\displaystyle T - 1 = \pm \sqrt{\frac{1}{5}} = \pm \frac{\sqrt{5}}{5}$

    $\displaystyle T = 1 \pm \frac{\sqrt{5}}{5}$

    -Dan
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  4. #4
    Forum Admin topsquark's Avatar
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    Quote Originally Posted by danielwu
    $\displaystyle (3)$In machine design. in finding the outside diameter$\displaystyle D_o$ of a hollow shaft, the equation$\displaystyle D_o^2 - DD_o - 0.25D^2 = 0$ is used Solve for $\displaystyle D_o$ if $\displaystyle D = 3.625cm$
    Same method. Get the constant term on the RHS of the equation, find out what "a" is, then add $\displaystyle a^2$ to both sides and go from there.

    -Dan
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by danielwu
    alright i have solve it.

    but here is another question which i also don't know how to do.
    can help me.

    ..[snip]
    Please post new unrelated questions in a new thread, it reduces confusion

    Thanks

    RonL

    (I will split this question out into a new thread)
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