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because 3*3=9 and not 6

x^2-9 = x^2-3^2 = (x+3)(x-3)

x^2-6 = x^2-9+3 = (x+3)(x-3)+3

typo its suppose to be (x-3) (x+2)

ohh,
(x-3)(x+2) = x^2 -3x+2x -6
=x^2 -x -6
and not x^2 -6

if factored is it (x-3) (x+2) or square root +-6?

roots are \(\large \pm \sqrt6\)
factors are \((x+\sqrt6)(x-\sqrt6) \)

but it is not (x-3) (x+2)

exatly! its not
because (x-3)(x+2) = x^2-x-6

ooh i see

(x-3) (x-2)= x^2-2x-3x+6=x^2-5x+6...so u got an extra 5x

ask if still any doubts.

so when ever it is not a perfect square, i have to to take the square root of both side?

You have to take the square root regardless.

whenever ?
when its in the form of
x^2 - a^2 = 0

then shift a^2 to other side and take square root.

k thanks

welcome :)

bcs u hv 1 exttra x :)

:D