• Aug 21st 2008, 10:23 AM
Peyton Sawyer
2y^2+2y+1=0
• Aug 21st 2008, 10:25 AM
Chop Suey

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

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

$2y^2+2y+1 = ay^2 + by + c$
• Aug 21st 2008, 10:29 AM
Chris L T521
Quote:

Originally Posted by Chop Suey

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

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

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

Quote:

Originally Posted by Peyton Sawyer
2y^2+2y+1=0

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

--Chris
• Aug 21st 2008, 10:31 AM
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 }
• Aug 21st 2008, 10:36 AM
Chris L T521
Quote:

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 }

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
• Aug 21st 2008, 10:41 AM
Peyton Sawyer
this is the original

2y^2+ 2y=1
• Aug 21st 2008, 10:43 AM
Chop Suey
In this case, you rearrange it to:
$2y^2 + 2y - 1 = 0$

And use the quadratic formula to find the two roots.
• Aug 21st 2008, 10:46 AM
mathceleb

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

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

-2 ± $\sqrt{12}$/4

Then Factor out a 2 from the top, which, for the second term is the simplifying the square root.
• Aug 21st 2008, 10:48 AM
Peyton Sawyer
okay i got

-1 plus/minus sq root 3/2
• Aug 21st 2008, 10:52 AM
Chop Suey
$\frac{-1\pm\sqrt{3}}{2}$
Then yes, you are right.