Adi3
  • Adi3
will medal Please help how to solve (x-3)^2 - (x+4)^2
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
Adi3
  • Adi3
@ganeshie8
Adi3
  • Adi3
@iambatman
ganeshie8
  • ganeshie8
wat do u want to solve?

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More answers

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

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