1. ## Rectangle?

Hi!
There's again something too difficult for me, I hope you can help me
You have a garden whose area is 18 sq ft and perimeter's 18 ft. It's length is x and width is y.

The first question says to write equations about this, and I think I've gone so far: a=x*y. p=2y+2x. Right? OK. Now comes the extremly difficult part:

1)Prove that (x+y)^2-4xy=(x-y)^2?
With the expand-and-solve thing I ended up with this: x^2+y^2-4xy=x^2+y^2. Is that right? And how can it help me prove that (x+y)^2-4xy=(x-y)^2?

2)Using the results of question 1, calculate (x-y)^2 and find x-y.

3)Using the results of question 2, calculate x and y.

2. Originally Posted by Catherine
Hi!
There's again something too difficult for me, I hope you can help me
You have a garden whose area is 18 sq ft and perimeter's 18 ft. It's length is x and width is y.

The first question says to write equations about this, and I think I've gone so far: a=x*y. p=2y+2x. Right? OK. Now comes the extremly difficult part:

1)Prove that (x+y)^2-4xy=(x-y)^2?
With the expand-and-solve thing I ended up with this: x^2+y^2-4xy=x^2+y^2. Is that right? And how can it help me prove that (x+y)^2-4xy=(x-y)^2?

2)Using the results of question 1, calculate (x-y)^2 and find x-y.

3)Using the results of question 2, calculate x and y.
$\displaystyle (x+y)^2-4xy$ expand the binomial and we get

$\displaystyle x^2+2xy+y^2-4xy=x^2-2xy-y^2$ now we factor

$\displaystyle x^2-2xy+y^2=(x-y)(x-y)=(x-y)^2$

I hope this get started.

Good luck

3. Originally Posted by TheEmptySet
$\displaystyle (x+y)^2-4xy$ expand the binomial and we get

$\displaystyle x^2+2xy+y^2-4xy=x^2-2xy-y^2$ now we factor

$\displaystyle x^2-2xy+y^2=(x-y)(x-y)=(x-y)^2$

I hope this get started.

Good luck
Thanks So for question n° 2 this means that (x-y)^2=x^2+y^2-2xy. But how can this help me find x-y and how can I calculate x and y using this? Thanks.

4. You know that 18 = 2x+2y which gives you (x+y) = 9. also xy = 18.

So (x-y)^2 = (x+y)^2 -4xy = 9^2 - 4*8 = 81-72 = 9

So (x-y) = 3

Solve for x and y.

5. Thanks! Now this is resolved!