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wat do u want to solve?
(x-3)^2 - (x+4)^2
solve for x or what
-7 (2x+1) this is the answer
we need to factor ths solution out to the answer igave u
but i dont know how to do tht
Ah, ok start of by expanding \[(x-3)^2 = (x-3)(x-3)\] and do the same with the other expression
You know how to distribute right?
Ok go ahead and try :)
x^2 + 9
@zzr0ck3r wht next
\[(x-3)^2 \neq x^2+9\]
\(a^2-b^2=(a-b)(a+b)\) Use that and we get \[(x-3)^2-(x-4)^2=[(x-3)+(x-4)][(x-3)-(x-4)]=(2x-7)(1)\\=2x-7\]
@iambatman isnt @zzr0ck3r doing the right thing
Sorry to but in @iambatman I just thought this might be the best way to go about it.
Don't worry about it! @Adi3 he's doing it right, just another method!
I suggest doing both @Adi3
the right answer is -7(2x + 1)
so how is @zzr0ck3r method correct
There was a sign error it's (x+4)^2 but otherwise he's right!
i m sorry but i still dont get it
Well lets do it both ways, so you can see, you can distribute and all that as I showed above or we can do it with zz's method \[a^2-b^2=(a-b)(a+b)\] here we let \[a^2 = (x-3)^2\] and \[b^2=(x+4)^2\] setting up we have \[(x-3)^2-(x+4)^2=[(x-3)+(x+4)][(x-3)-(x+4)]=-7(2x+1)\]
ohh now i get it
thnxs who should i medal
i cant medal u both though i want to
so who should i medal
yep u r right everyone got a medal except @ganeshie8
@Adi3 you can medal both if you like both answers, there is a way to do it but it is not official yet
alright thnxs everyone