Question

will medal
Please help
how to solve (x-3)^2 - (x+4)^2

Answer 1

@ganeshie8

Answer 2

@iambatman

Answer 3

wat do u want to solve?

Answer 4

(x-3)^2 - (x+4)^2

Answer 5

solve for x or what

Answer 6

-7 (2x+1) this is the answer

Answer 7

we need to factor ths solution out to the answer igave u

Answer 8

but i dont know how to do tht

Answer 9

Ah, ok start of by expanding \[(x-3)^2 = (x-3)(x-3)\]
and do the same with the other expression

Answer 10

You know how to distribute right?

Answer 11

yes

Answer 12

Ok go ahead and try :)

Answer 13

\(a^2-b^2=(a-b)(a+b)\)

Answer 14

x^2 + 9

Answer 15

@zzr0ck3r wht next

Answer 16

\[(x-3)^2 \neq x^2+9\]

Answer 17

\(a^2-b^2=(a-b)(a+b)\)

Use that and we get

\[(x-3)^2-(x-4)^2=[(x-3)+(x-4)][(x-3)-(x-4)]=(2x-7)(1)\\=2x-7\]

Answer 18

@iambatman isnt @zzr0ck3r doing the right thing

Answer 19

Sorry to but in @iambatman I just thought this might be the best way to go about it.

Answer 20

Don't worry about it! @Adi3 he's doing it right, just another method!

Answer 21

I suggest doing both @Adi3

Answer 22

the right answer is -7(2x + 1)

Answer 23

so how is @zzr0ck3r method correct

Answer 24

There was a sign error it's (x+4)^2 but otherwise he's right!

Answer 25

i m sorry but i still dont get it

Answer 26

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 \[(x-3)^2-(x+4)^2=[(x-3)+(x+4)][(x-3)-(x+4)]=-7(2x+1)\]

Answer 27

ohh now i get it

Answer 28

thnxs who should i medal

Answer 29

i cant medal u both though i want to

Answer 30

so who should i medal

Answer 31

Ganeshie!

Answer 32

yep u r right everyone got a medal except @ganeshie8

Answer 33

Hehe

Answer 34

@Adi3 you can medal both if you like both answers, there is a way to do it but it is not official yet

Answer 35

nice

Answer 36

alright thnxs everyone