Trouble Solving quadratic: 1 = (y + 3)(2y - 2)

[Jen123]

Solve 1 = (y + 3)(2y - 2)

What I tried:
I always come to a freezing point no matter what way i try this problem
first i tried completeing the square method but that didn't work I also tried factoring but i couldnt figure it out. How would I begin and solve this problem??

 

[Deleted member 4993]

Re: Trouble Solving

Solve 1=(y+3)(2y-2)

What I tried:
I always come to a freezing point no matter what way i try this problem
first i tried completeing the square method but that didn't work I also tried factoring but i couldnt figure it out. How would I begin and solve this problem??

First

Use FOIL to distrubute and expand the function:

(y+3)(2y-2) = 1

2y^2 + 4y - 6 = 1

Simplify

2y^2 + 4y - 7 = 0

Now factorize the quadratic equation with your favorite method and solve for y.

 

[Jen123]

Yes that is where i get stuck at when i try to factor. I didvide the equation by 2 to get y^2 by itself: y^2+2y-7/2
I dont know what will multiply to get 7/2 and add to get 2, so what should i do??

 

[jwpaine]

You should complete the square (quadratic formula) to find the two roots - this trinomial does not have rational roots

 

[Jen123]

I tried that to and I got:
2(y+1)^2=9
So how would I write the two equations to set them equal to zero so i can solve for y?

 

[jwpaine]

I tried that to and I got:
2(y+1)^2=9
So how would I write the two equations to set them equal to zero so i can solve for y?

Not sure what you mean by 'two equations'

You should complete the square: http://www.purplemath.com/modules/sqrquad.htm

 

[skeeter]

I tried that to and I got:
2(y+1)^2=9
So how would I write the two equations to set them equal to zero so i can solve for y?

you completed the square correctly, proceed to solve for y as follows ...

(y + 1)² = 9/2

y + 1 = +/- sqrt{9/2}

y = -1 +/- sqrt{9/2}