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OpenStudy (anonymous):
a)1
b)6
c)4
d)-1
OpenStudy (anonymous):
1
OpenStudy (anonymous):
you sure?
OpenStudy (anonymous):
@03453660
OpenStudy (anonymous):
yap
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OpenStudy (anonymous):
want to know??
OpenStudy (anonymous):
sure
OpenStudy (anonymous):
@03453660
OpenStudy (anonymous):
ok
OpenStudy (anonymous):
your equation is
x^2-2x+6=0
now take the constant to RHS
x^2-2x = -6
look to LHS of this equation what should i add to make it perfect square like it become (a+b)^2 or (a-b)^2
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OpenStudy (anonymous):
ok
OpenStudy (anonymous):
so if i add 1 to LHS and RHS then LHS become a perfect square
x^2-2x+1 = -6+1
(x-1)^2 = -5
this LHS has became a perfect square by adding 1 to it
OpenStudy (anonymous):
got it!! thanks!!
OpenStudy (anonymous):
welcome! and sorry for the late reply i have some problem in internet connection
OpenStudy (anonymous):
no worries..could you help me with one more please? x^2-4x-9=0 @03453660
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OpenStudy (anonymous):
a)9
b)-4
c)4
d)13
OpenStudy (anonymous):
this is 4
OpenStudy (anonymous):
did you know how to solve this. its a little bit tricky