Solve for the variable: 2(y^2 - 2y + 1) = 4(y^2 + 2y + 1)

[Guest]

I need help because I keep getting stuck.

2(y^2 - 2y + 1) = 4(y^2 + 2y + 1)
2y^2 - 4y + 2 = 4y^2 + 8y + 4
2y^2 - 4y^2 - 4y - 8y + 2 - 4 = 0
(-2y^2 - 12y - 2) = 0
-2(y^2 + 6y + 1)

Next I tried to factor but it doesn't work.
Thanks in advance for any help.

Andrea

 

[skeeter]

Re: Solve for the variable

I need help because I keep getting stuck.

2(y^2 - 2y + 1) = 4(y^2 + 2y + 1)
2y^2 - 4y + 2 = 4y^2 + 8y + 4
2y^2 - 4y^2 - 4y - 8y + 2 - 4 = 0
(-2y^2 - 12y - 2) = 0
-2(y^2 + 6y + 1)

Next I tried to factor but it doesn't work.
Thanks in advance for any help.

Andrea

you're correct, the quadratic will not factor. have you seen this before?

\(\displaystyle \L
\frac{-b \pm \sqrt{b^2 - 4ac}}{2a}\)

 

[Guest]

Re: Solve for the variable

So I correct then and this is the final answer?

Thanks!!
Andrea

 

[stapel]

Re: Solve for the variable

So I [was] correct then...?

You were correct that the quadratic does not factor. But "it doesn't factor" is not the answer. You need to apply the Quadratic Formula (as provided earlier) to find the solution.

Eliz.

 

[Guest]

Solve for the variable

Omg I cannot figure out how to paste my work here so it looks right. I used the quadratic and ended up with an answer of 3 +/- sqrt (12^2-48). Would that be correct? I can post my work if someone can tell me how.

Thanks!!
Andrea

 

[stapel]

Re: Solve for the variable

...ended up with an answer of 3 +/- sqrt (12^2-48).

Sorry, but no.

As for formatting, either use LaTeX (explained in the links in the "Forum Help" pull-down menu at the very top of this page), or else just use grouping symbols. For instance, you started with a = 1, b = 6, c = 1, so:

. . . . .y = [-(6) ± sqrt((6)^2 - 4(1)(1))] / [2(1)]

Then what?

Eliz.

 

[Guest]

okay hopefully you can understand this cuz I don't have time to try to figure out LaTeX tonite.

2(y^2 - 2y + 1) = 4(y^2 + 2y + 1)
2y^2 - 4y + 2 = 4y^2 + 8y + 4
2y^2 - 4y^2 - 4y - 8y + 2 - 4 = 0
-2y^2 - 12y - 2 = 0

so I used
a = -2
b = -12
c = -2

12 +/- sqrt[-12^2 - (4)(-2)(-2)] / [(2)(-2)]
12 +/- sqrt[144 - 16] / -4
-3 +/- sqrt 128

Make sense? I have it in a Word Doc using Math Type.

 

[skeeter]

why are you making it difficult ?

-2y^2 - 12y - 2 = 0
simplify by dividing each term by -2 ...
y^2 + 6y + 1 = 0

a = 1, b = 6, c = 1

\(\displaystyle y = \frac{-6 \pm \sqrt{6^2 - 4(1)(1)}}{2(1)}\)

can you take it from here?

 

[Denis]

Re: Solve for the variable: 2(y^2 - 2y + 1) = 4(y^2 + 2y + 1

2(y^2 - 2y + 1) = 4(y^2 + 2y + 1)

Quicker and simpler still; divide by 2 right away:
y^2 - 2y + 1 = 2(y^2 + 2y + 1)

2y^2 + 4y + 2 = y^2 - 2y + 1

2y^2 - y^2 + 4y + 2y + 2 - 1 = 0

y^2 + 6y + 1 = 0

now call on ye olde quadratic :wink:

You did this on your attempt:
y = 12 +/- sqrt[144 - 16] / -4
y = -3 +/- sqrt(128)

That's a no-no; should be y = -3 +/- sqrt(128) / -4
And you needed another set of brackets in your 1st line:
y = [12 +/- sqrt[144 - 16]] / -4

Btw, sqrt(128) = sqrt(64 * 2) = 8 * sqrt(2) : or 8sqrt(2)