if x=3 is one root of the quadratic equation x2-2kx-6=0,then find the value of k
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Answered by
2
here is your answer by Sujeet,
x²-2kx-6=0
3²-2k(3)-6=0
9-6k-6=0
-6k=6-9
-6k=-3
k=3/6
k=1/2
that's all Sujeet...
Follow me.
x²-2kx-6=0
3²-2k(3)-6=0
9-6k-6=0
-6k=6-9
-6k=-3
k=3/6
k=1/2
that's all Sujeet...
Follow me.
Answered by
1
if x=3 then
(3)×2-2k(3)-6=0
6-6k-6=0
6-6=6k
6k=0
k=0/6
therefore k=0
(3)×2-2k(3)-6=0
6-6k-6=0
6-6=6k
6k=0
k=0/6
therefore k=0
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