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Math Help - Completing the square

  1. #1
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    Completing the square

    Can anyone solve this quadratic by completing the square for me?
    <br />
16y^2+16y=1<br />
    Steps you took?
    Last edited by EyesForEars; October 24th 2008 at 05:35 AM.
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  2. #2
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    Solution

    It's been a few years since I've had a problem like this, so hopefully I am going about this correctly. Start by dividing the equation through by 16 to get:

    y^2 + y = 1/16

    Now, we figure that in order to factor the left hand side of the equation we would need something of the form (y + c)^2 with c denoting some constant value.

    We would like to choose something along the lines of:

    (y+1/2)^2 because when you multiply it out you get:

    y^2 + y + 1/4. But since we have this additional 1/4 on the left side of the equation, we must add it to the right side to get the full equation:

    y^2 + y + 1/4 = 1/16 + 1/4

    This becomes: (y + 1/2)^2 = (1/4) + (1/16)
    which becomes: (y + 1/2)^2 = 5/16.

    Now take the square root of both sides to yield:

    y + 1/2 = +/- sqrt(5)/4

    Final step: subtract 1/2 from both sides to solve for y:

    y = (+/- sqrt(5)/4) - 1/2

    I hope this explanation is useful.
    Last edited by ajj86; October 24th 2008 at 05:42 AM. Reason: remove extra steps
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  3. #3
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    Most. Thanx a bunch.
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  4. #4
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    Quote Originally Posted by EyesForEars View Post
    Can anyone solve this quadratic by completeing the square for me?
    <br />
16y^2+16y=1<br />
    Steps you took?
    Edit: This is late, but here it is anyway.

    First thing, we have to make the coefficient of the quadratic term 1. So divide all terms by 16.

    16y^2+16y=1

    y^2+y=\frac{1}{16}

    Now, take half of the coefficient of the linear term, square it, and add it to both sides.

    y^2+y+\frac{1}{4}=\frac{1}{16}+\frac{1}{4}

    (y+\frac{1}{4})^2=\frac{5}{16}

    Now, take the square root of both sides.

    y+\frac{1}{2}=\frac{\pm \sqrt{5}}{4}

    y=\frac{\pm \sqrt{5}}{4}-\frac{1}{2}=\frac{\pm \sqrt{5}-2}{4}
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