SOLUTION: Solve for the variable: 2(y � 1)^2 = 4(y + 1)^2 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 -2y^2 � 12y � 2 = 0 a = -2 b

Question 49645: Solve for the variable:
2(y � 1)^2 = 4(y + 1)^2
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
-2y^2 � 12y � 2 = 0
a = -2
b = -12
c = -2
2 +/- 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.

Answer by gsmani_iyer(201) (Show Source): You can put this solution on YOUR website!

Hello,
You have done correctly till the third line from the last. So your answer has gone wrong. From the second line from the last I have done it here. Please check up.
=
Now let us take

= = ... ... (1)

Now let us take

= = ..... .... (2)
From the (1) and (2) above, we can make out
Solution set = (, ) Answer.
I hope this is clear to you.
RELATED QUESTIONS
I am having some troubles with solving for the variable in equation... (answered by josmiceli)

2/y+1=-4y/2y+2+3 (answered by Alan3354)

I'm stumped. Solve for Y 4(2y-5)=2(4y-10) 8y-20=8y-20 8y-20+20=8y-20+20 8y=8y... (answered by venugopalramana)

Can you solve y^2-2y-8x-16=0 and x^2+4y^2-8y=4? (answered by jsmallt9)

Could you please help me find the inverse of the following function? f(x) =... (answered by lwsshak3)

solve by comparison x+2y =4 x+y=2 another x-4y=1... (answered by Theo)

Multiply {{{... (answered by mananth)

8y^2-8y-4/2y-1 (answered by ikleyn)

what are the factors of the following trinomial? 16y^2 - 12y + 2 a)(2y - 1) (8y +... (answered by Alan3354)