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Thread: Basic polynomial

  1. #1
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    Basic polynomial

    I have a eq $\displaystyle 2y^2 + y + 1 = 0$

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

    $\displaystyle 2y^2 + y -1 +2 = 0$

    $\displaystyle 2y^2 + y -1 = -2$

    $\displaystyle (2y-1)(y+1) = -2$

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

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

    & similarly

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

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

    $\displaystyle 2y^2 + y -1 +2 = 0$

    $\displaystyle 2y^2 + y -1 = -2$

    $\displaystyle (2y-1)(y+1) = -2$

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

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

    & similarly

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

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


    If you check the discriminant

    $\displaystyle \Delta = b^2 - 4ac$

    $\displaystyle = 1^2 - 4(1)(1)$

    $\displaystyle = 1 - 4$

    $\displaystyle = -3$

    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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    Quote Originally Posted by wolfyparadise View Post
    So what is the value of 'y'
    $\displaystyle y = \frac{-1 \pm i\sqrt{7}}{4}$
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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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