If x=3 is one of the root of the equation x2_2kx_6=0, find the value of k?
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Solution==>
p(x)=x²-2kx-6=0
given x=3
so, put values of x in p(x)
(3)²-2k(3)-6=0
9-6k-6=0
-6k=-9+6
-6k=-3
k=-3/-6
k=1/2
•.• value of k is =1/2
p(x)=x²-2kx-6=0
given x=3
so, put values of x in p(x)
(3)²-2k(3)-6=0
9-6k-6=0
-6k=-9+6
-6k=-3
k=-3/-6
k=1/2
•.• value of k is =1/2
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