# So, a confusion on a^2-b^2 sums. Any explanation? c:

• Mar 30th 2013, 04:54 AM
So, a confusion on a^2-b^2 sums. Any explanation? c:
It's not actually a problem related to a question, it's a doubt that has been nagging my mind since yesterday ^^
So, why is
............................. $a^{2}-b^{2}=\left ( a-b \right )\left ( a+b \right )$?

Can't it be that
............................. $a^{2}-b^{2}=\left ( a-b \right )^{2}$

?

And yes, when expanding
............................. $a^{2}-b^{2}=\left ( a-b \right )\left ( a+b \right )$
............................. $a^{2}-b^{2}=a\left ( a-b \right )-b\left ( a-b \right )$
............................. $a^{2}-b^{2}=a^{2}+ab-ab-b^{2}$
............................. $a^{2}-b^{2}=a^{2}-b^{2}$

and when expanding
............................. $a^{2}-b^{2}=\left ( a-b \right )^{2}$
............................. $a^{2}-b^{2}=a^{2}-2ab+b^{2}$
............................. $a^{2}-b^{2}=a^{2}-ab-ab+b^{2}$
............................. $a^{2}-b^{2}=a\left ( a-b \right )-b\left ( a-b \right )$
............................. $a^{2}-b^{2}=\left ( a-b \right )\left ( a-b \right )$
............................. $a^{2}-b^{2}=\left ( a-b \right )^{2}$

Can't $a^{2}-b^{2}= \left ( a-b \right )\left ( a+b \right )=\left ( a-b \right )^{2}$?
• Mar 30th 2013, 04:59 AM
HallsofIvy
Re: So, a confusion on a^2-b^2 sums. Any explanation? c:
If a= 5 and b= 3 then a- b= 2 and a+ b= 8. (a- b)(a+ b)= 2(8)= 16 while (a- b)^2= 4. Now, what is a^2- b^2?
• Mar 30th 2013, 04:59 AM
mrmaaza123
Re: So, a confusion on a^2-b^2 sums. Any explanation? c:
Well just solve (a-b)(a+b) and (a-b)(a-b) and you will see the difference yourself.
• Mar 30th 2013, 05:07 AM
Re: So, a confusion on a^2-b^2 sums. Any explanation? c:
Quote:

Originally Posted by HallsofIvy
If a= 5 and b= 3 then a- b= 2 and a+ b= 8. (a- b)(a+ b)= 2(8)= 16 while (a- b)^2= 4. Now, what is a^2- b^2?

Oh I understand. So basically,
$\left ( a-b \right )\left ( a+b \right )=16$
$\left ( a-b \right )^{2} = 4$
$a^{2}-b^{2} = 5^{2}-3^{2}$
$a^{2}-b^{2} = 25 - 9$
$a^{2}-b^{2} = 16$

So finally, $\left ( a-b \right )\left ( a+b \right )=\left ( a-b \right )^{2}$
?
• Mar 30th 2013, 05:13 AM
MINOANMAN
Re: So, a confusion on a^2-b^2 sums. Any explanation? c:
Nnnnnnnnnnnnnnnnnnnnnnoooooooooooooooooooooooooooo ooooooo
• Mar 30th 2013, 05:43 AM
Re: So, a confusion on a^2-b^2 sums. Any explanation? c:
Quote:

Originally Posted by MINOANMAN
Nnnnnnnnnnnnnnnnnnnnnnoooooooooooooooooooooooooooo ooooooo

I see. Then no it may be :)
• Mar 30th 2013, 05:49 AM
emakarov
Re: So, a confusion on a^2-b^2 sums. Any explanation? c:
Quote:

Originally Posted by Lexadis
$\left ( a-b \right )\left ( a+b \right )=16$
$\left ( a-b \right )^{2} = 4$
...
So finally, $\left ( a-b \right )\left ( a+b \right )=\left ( a-b \right )^{2}$
?

Yes. So, you wouldn't mind then if you give me $16 and I will give you back the same$4? Then we'll be even.
• Mar 30th 2013, 06:00 AM
Soroban
Re: So, a confusion on a^2-b^2 sums. Any explanation? c:

You're making silly errors . . .

Quote:

Why is . $a^2-b^2\:=\:( a-b)(a+b )\,?$

Can't it be that: . $a^2-b^2\:=\: (a-b)^2$ . No!

When expanding
. . $a^{2}-b^{2}\:=\:(a-b)(a+b)$
. . $a^2-b^2 \:=\:a (a-b)-b(a-b)$ . ${\color{blue}\text{This is }(a-b)(a-b)}$
. . $a^2-b^2 \:=\:a^2\,{\color{red}+}\,ab-ab-b^2$
. . $a^2-b^2 \:=\:a^2-b^2$

You managed to come up with the correct result
. . after making two errors.

Quote:

When expanding
. . $a^{2}-b^{2}\:=\:\left ( a-b \right )^{2}$ . . . . . . If this is true,
. . $a^{2}-b^{2}\:=\:a^{2}-2ab+b^{2}$
. . $a^{2}-b^{2}\:=\:a^{2}-ab-ab+b^{2}$
. . $a^{2}-b^{2}\:=\:a\left ( a-b \right )-b\left ( a-b \right )$
. . $a^{2}-b^{2}\:=\:\left ( a-b \right )\left ( a-b \right )$
. . $a^{2}-b^{2}\:=\:\left ( a-b \right )^{2}$ . . . . . . then this is true.

That was a waste of time . . .

You began with an equation which may or may not be true.

You expanded the right hand side.

Then you factored the right hand side.

And you returned to the equation which may or may not be true.

I can do that, too.

. . $13 \:=\:30$

. . $13 \:=\:5\cdot6$

. . $13 \:=\:5\cdot2\cdot3$

. . $13 \:=\:10\cdot3$

. . $13 \:=\:30$

Quote:

$\text{Can't }a^{2}-b^{2}\:=\: (a-b)(a+b)\:=\: (a-b)^2\,?$

Do you realize what you said?

. . $(a-b)(a+b) \:=\:(a-b)^2$

. . $(a-b)(a+b) \:=\:(a-b)(a-b)$

. . $\frac{(a-b)(a+b)}{a-b} \:=\:\frac{(a-b)(a-b)}{a-b}$

. . . . . . . . $a+b \:=\:a-b$

$\text{So adding }a\text{ and }b\text{ is the same as subtracting }a\text{ and }b\,!$
• Mar 30th 2013, 08:25 AM
Re: So, a confusion on a^2-b^2 sums. Any explanation? c:
Quote:

Originally Posted by Soroban

You're making silly errors . . .

You managed to come up with the correct result
. . after making two errors.

That was a waste of time . . .

You began with an equation which may or may not be true.

You expanded the right hand side.

Then you factored the right hand side.

And you returned to the equation which may or may not be true.

I can do that, too.

. . $13 \:=\:30$

. . $13 \:=\:5\cdot6$

. . $13 \:=\:5\cdot2\cdot3$

. . $13 \:=\:10\cdot3$

. . $13 \:=\:30$

Do you realize what you said?

. . $(a-b)(a+b) \:=\:(a-b)^2$

. . $(a-b)(a+b) \:=\:(a-b)(a-b)$

. . $\frac{(a-b)(a+b)}{a-b} \:=\:\frac{(a-b)(a-b)}{a-b}$

. . . . . . . . $a+b \:=\:a-b$

$\text{So adding }a\text{ and }b\text{ is the same as subtracting }a\text{ and }b\,!$

Didn't think of it that way.. and thank you, got it :)