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

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 $y$

$$6y-4 = 4$$

$$6y = 8$$

But the answer key has $y=4$

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$y=4$ is clearly a solution since $4\times 9 = 6^2$ – Henry Oct 31 '11 at 18:22
Essentially (as others have pointed out), your error lies in the idea that $(a +b)^2 = a^2 +b^2$, which is almost always false. You can see this quickly by trying $a=b=1$. Then you have $(1+1)^2 ?= 1^2 +1^2$ – The Chaz 2.0 Oct 31 '11 at 18:34
There is still something wrong at combine $y$ on the left side, can't figure out what. – Liger86 Oct 31 '11 at 18:45

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 $(2y-2)(2y-2)$ – Liger86 Oct 31 '11 at 18:21

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 $y=4$.

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