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I'm getting different answer from answer key.

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

FOIL left side $4y^2+2y-8y-4 = (2y-2)^2$

Right side $4y^2+2y-8y-4 = 4y^2+4 $

Subtract $4y^2$ from both sides

$2y-8y-4 = 4 $

Combine $y$

$6y-4 = 4$

add 4 to both sides

$6y = 8$

But the answer key has $y=4$

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    There is still something wrong at combine $y$ on the left side, can't figure out what.2011-10-31

2 Answers 2

4

You're unfolding the right-hand side wrong -- $(2y-2)^2$ is not $4y^2+4$, but $4y^2+4-8y$.

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    Ah, I see it's $(2y-2)(2y-2)$2011-10-31
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The error is in the "Right Side" step.

You essentially wrote $(2y-2)^2 = 4y^2 + 4.$ That's incorrect.

Remember: $(a-b)^2 = a^2 - 2ab + b^2$. So $(2y-2)^2 = 4y^2 - 8y + 4.$

The third displayed equation should thus be $4y^2 +2y - 8y - 4 = 4y^2 -8y + 4.$ You will find that this leads to $2y = 8$, from which you get $y=4$.