Basic polynomial

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December 27th 2009, 05:46 PM
wolfyparadise

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

$y = -2-1 = -3$
December 27th 2009, 05:57 PM
Prove It

Quote:

Originally Posted by wolfyparadise

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.
December 27th 2009, 06:42 PM
wolfyparadise

So what is the value of 'y'
December 27th 2009, 07:08 PM
skeeter

Quote:

Originally Posted by wolfyparadise

So what is the value of 'y'

$y = \frac{-1 \pm i\sqrt{7}}{4}$
December 27th 2009, 07:14 PM
Prove It

Quote:

Originally Posted by wolfyparadise

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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