quadratic formula

2y^2+2y+1=0

The quadratic formula is:

$\displaystyle y = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

What are a, b, and c? Well, you should know that:

$\displaystyle 2y^2+2y+1 = ay^2 + by + c$

Quote:

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Originally Posted by Chop Suey

The quadratic formula is:

$\displaystyle y = \frac{-b\pm\sqrt{b^2-4ac}}{2a}$

What are a, b, and c? Well, you should know that:

$\displaystyle 2y^2+2y+1 = ay^2 + by + c$

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Quote:

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Originally Posted by Peyton Sawyer

2y^2+2y+1=0

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Just keep in mind that you will get complex [imaginary] solutions since the discriminant $\displaystyle b^2-4ac<0$. Note that $\displaystyle \sqrt{-1}=i$

--Chris

my answers arnt even close to these


a. {(± √3)/2}


b. { -1, 0 }


c. {(1 ± √3)/2}


d. {(-1 ± √3)/2}


e. { 0, 1 }

Quote:

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Originally Posted by Peyton Sawyer

my answers arnt even close to these


a. {(± √3)/2}


b. { -1, 0 }


c. {(1 ± √3)/2}


d. {(-1 ± √3)/2}


e. { 0, 1 }

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Just check and make sure you typed out the equation correctly. I think you have a missing minus sign somewhere. The way you gave us the equation implied that there was no real solution. If this equation you gave us is correct, then none of these choices are correct.

--Chris

this is the original

2y^2+ 2y=1

In this case, you rearrange it to:
$\displaystyle 2y^2 + 2y - 1 = 0$

And use the quadratic formula to find the two roots.

Answer is d.

I need to build an extra piece on the lesson here:

Quadratic Equation

But you can follow the math 1/2 way down and see that you get to:

-2 ± $\displaystyle \sqrt{12}$/4

Then Factor out a 2 from the top, which, for the second term is the simplifying the square root.

okay i got

-1 plus/minus sq root 3/2

$\displaystyle \frac{-1\pm\sqrt{3}}{2}$
Then yes, you are right.