1. ## 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

$\displaystyle y = -2-1 = -3$

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.

3. So what is the value of 'y'

$\displaystyle y = \frac{-1 \pm i\sqrt{7}}{4}$