Question 117427
{{{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}}}
Subtract {{{2y^2}}} from both sides
{{{-4y + 2 = 2y^2 + 8y + 4}}}
Add {{{4y}}} to both sides
{{{2 = 2y^2 + 12y + 4}}}
Subtract {{{2}}} from both sides
{{{2y^2 + 12y + 2 = 0}}}
divide both sides by {{{2}}}
{{{y^2 + 6y + 1 = 0}}}
Now subtract {{{1}}} from both sides
{{{y^2 + 6y = -1}}}
Now complete the square by taking 1/2 of the 
coefficient of y, square it, and add it to both sides
{{{y^2 + 6y + 9 = -1 + 9}}}
{{{(y + 3)^2 = 8}}}
Take the square root of both sides
{{{y + 3 = 0 +- sqrt(8)}}}
{{{y = -3 +- 2*sqrt(2)}}}
{{{y = -3 + 2*sqrt(2)}}} answer
{{{y = -3 - 2*sqrt(2)}}} answer
check answer
{{{2(y-1)^2 = 4(y+1)^2}}}
{{{(-3 + 2*sqrt(2) - 1)^2 = 2(-3 + 2*sqrt(2) + 1)^2}}}
{{{16 - 16*sqrt(2) + 8 = 2(4 - 8*sqrt(2) + 8)}}}
{{{16 - 16*sqrt(2) + 8 = 8 - 16*sqrt(2) + 16}}}
{{{24 - 16*sqrt(2) = 24 - 16*sqrt(2)}}}
{{{3 - 2*sqrt(2) = 3 - 2*sqrt(2)}}}
OK