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

1. ## 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 $a^{2}-b^{2}= \left ( a-b \right )\left ( a+b \right )=\left ( a-b \right )^{2}$?

2. ## 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?

3. ## 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.

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

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, $\left ( a-b \right )\left ( a+b \right )=\left ( a-b \right )^{2}$
?

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

Nnnnnnnnnnnnnnnnnnnnnnoooooooooooooooooooooooooooo ooooooo

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

Originally Posted by MINOANMAN
Nnnnnnnnnnnnnnnnnnnnnnoooooooooooooooooooooooooooo ooooooo
I see. Then no it may be

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

$\left ( a-b \right )\left ( a+b \right )=16$
$\left ( a-b \right )^{2} = 4$
...
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.

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

You're making silly errors . . .

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.

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.

. . $13 \:=\:30$

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

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

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

. . $13 \:=\:30$

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

Do you realize what you said?

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

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

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

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

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

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

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" alt="(a-b)(a+b) \:=\a-b)^2" />

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