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

Algebra ->  Coordinate Systems and Linear Equations -> 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       Log On


   

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) About Me  (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.
%2812%2B-sqrt%28128%29%29%2F%28-4%29%29 = %2812+%2B-+8sqrt%282%29%29%2F%28-4%29%29
Now let us take %2812+%2B+8sqrt%282%29%29%2F%28-4%29%29

%2812+%2B+8sqrt%282%29%29%2F%28-4%29%29 = -3%2B%28-2%29sqrt%282%29 = -3-2sqrt%282%29 ... ... (1)

Now let us take %2812+-+8sqrt%282%29%29%2F%28-4%29%29

%2812+-+8sqrt%282%29%29%2F%28-4%29%29 = -3-%28-2%29sqrt%282%29 = -3%2B2sqrt%282%29 ..... .... (2)
From the (1) and (2) above, we can make out
Solution set = (-3+-2sqrt%282%29, -3%2B2sqrt%282%29) Answer.
I hope this is clear to you.