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Math Help - Basic polynomial

  1. #1
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    Basic polynomial

    I have a eq 2y^2 + y + 1 = 0

    Now what i did was 2y^2 + y + 1 -2 + 2 = 0

    2y^2 + y -1 +2 = 0

    2y^2 + y -1  = -2

    (2y-1)(y+1) = -2

    \therefore since (2y - 1) = -2

    y = \frac{-2 +1}{2} ..........Is this correct ?

    & similarly

    y = -2-1 = -3
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  2. #2
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    Quote Originally Posted by wolfyparadise View Post
    I have a eq 2y^2 + y + 1 = 0

    Now what i did was 2y^2 + y + 1 -2 + 2 = 0

    2y^2 + y -1 +2 = 0

    2y^2 + y -1 = -2

    (2y-1)(y+1) = -2

    \therefore since (2y - 1) = -2

    y = \frac{-2 +1}{2} ..........Is this correct ?

    & similarly

    y = -2-1 = -3
    No it's not.

    For any quadratic ax^2 + bx + c if you are going to complete the square, you need to make sure that the a value is 1. Also if you are going to use the Null Factor Law, one side of the equation has to be 0.


    If you check the discriminant

    \Delta = b^2 - 4ac

     = 1^2 - 4(1)(1)

     = 1 - 4

     = -3

    you will see that there are not any real solutions.
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  3. #3
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    So what is the value of 'y'
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  4. #4
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    Quote Originally Posted by wolfyparadise View Post
    So what is the value of 'y'
    y = \frac{-1 \pm i\sqrt{7}}{4}
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  5. #5
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    Quote Originally Posted by wolfyparadise View Post
    So what is the value of 'y'
    I just told you, there are not any real solutions.

    If you are allowed complex solutions, see Skeeter's post.
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