Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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 $y=4$

share|improve this question
    
$y=4$ is clearly a solution since $4\times 9 = 6^2$ –  Henry Oct 31 '11 at 18:22
1  
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

2 Answers 2

up vote 4 down vote accepted

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

share|improve this answer
    
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 $y=4$.

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.