# help please-hard question

• Mar 25th 2013, 11:24 PM
mathkid182
If 20 × 20 = 400, execute the difference of two squares to find 19×21?

I have no idea how to do this. any help would be great please?
• Mar 25th 2013, 11:40 PM
Gusbob
Re: help please-hard question
Quote:

Originally Posted by mathkid182
If 20 × 20 = 400, execute the difference of two squares to find 19×21?

I have no idea how to do this. any help would be great please?

The difference of two squares is $x^2-y^2=(x-y)(x+y)$. Let $x=20$ and $y=1$. Do you see where to go from here?
• Mar 26th 2013, 12:03 AM
mathkid182
Re: help please-hard question
not really im sorry, where did the 20 and the 1 come from? and what would be the next step?
would it be

20^2 - 1^2 = (20-1)(20+1)
400 - 1= 400+20-20-1
399 = 399
?thanks
• Mar 26th 2013, 12:12 AM
Gusbob
Re: help please-hard question
Quote:

Originally Posted by mathkid182
not really im sorry, where did the 20 and the 1 come from? and what would be the next step?
would it be

20^2 - 1^2 = (20-1)(20+1)
400 - 1= 400+20-20-1
399 = 399
?thanks

The 20 and the 1 basically came from the question. The point being $400-1=x^2-y^2=(x-y)(x+y)=(20-1)(20+1)= 19 \times 21$.
• Mar 26th 2013, 12:19 AM
mathkid182
Re: help please-hard question
Oh i think i get it now thank you so much. do think it would be fine to write the answer as:

=x2 - y2 = (x-y)(x+y)
=20^2 - 1^1 = (20-1)(20+1)
=400-1 = (20-1)(20+1)
=399 = 399

therefore 19x21 = 399

and that would be right?
• Mar 26th 2013, 12:30 AM
Gusbob
Re: help please-hard question
I would replace that last line with 399 = 19 x21 instead of 399 = 399. The latter does not actually constitute a proof.
• Mar 26th 2013, 12:32 AM
mathkid182
Re: help please-hard question
perfect thanks again your a great teacher!!