anonymous
  • anonymous
x^2-6=0 why isnt it (x+2) (x-3)=0 just need to know why
Mathematics
jamiebookeater
  • jamiebookeater
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hartnn
  • hartnn
because 3*3=9 and not 6
hartnn
  • hartnn
x^2-9 = x^2-3^2 = (x+3)(x-3)
hartnn
  • hartnn
x^2-6 = x^2-9+3 = (x+3)(x-3)+3

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anonymous
  • anonymous
typo its suppose to be (x-3) (x+2)
hartnn
  • hartnn
ohh, (x-3)(x+2) = x^2 -3x+2x -6 =x^2 -x -6 and not x^2 -6
anonymous
  • anonymous
if factored is it (x-3) (x+2) or square root +-6?
hartnn
  • hartnn
roots are \(\large \pm \sqrt6\) factors are \((x+\sqrt6)(x-\sqrt6) \)
anonymous
  • anonymous
but it is not (x-3) (x+2)
hartnn
  • hartnn
exatly! its not because (x-3)(x+2) = x^2-x-6
anonymous
  • anonymous
ooh i see
anonymous
  • anonymous
(x-3) (x-2)= x^2-2x-3x+6=x^2-5x+6...so u got an extra 5x
hartnn
  • hartnn
ask if still any doubts.
anonymous
  • anonymous
so when ever it is not a perfect square, i have to to take the square root of both side?
anonymous
  • anonymous
You have to take the square root regardless.
hartnn
  • hartnn
whenever ? when its in the form of x^2 - a^2 = 0
hartnn
  • hartnn
then shift a^2 to other side and take square root.
anonymous
  • anonymous
k thanks
hartnn
  • hartnn
welcome :)
anonymous
  • anonymous
bcs u hv 1 exttra x :)
anonymous
  • anonymous
:D

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