if x=3 is one root of the q.e x2-2kx-6=0,then find te value of k
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Answered by
3
hello friend your answer is
x=3
x²-2kx-6
3²-2k×3-6=0
9-6k-6=0
3-6k=0
k=1/2
x=3
x²-2kx-6
3²-2k×3-6=0
9-6k-6=0
3-6k=0
k=1/2
abhi569:
Nice
Answered by
2
x² -2kx -6=0
Put the value of x,
3² -2k*6 -6=0
9-6k -6 =0
9-6-6k=0
3-6k=0
3=6k
3/6 =k
1/2 =k
I hope this will help you
-by ABHAY
Put the value of x,
3² -2k*6 -6=0
9-6k -6 =0
9-6-6k=0
3-6k=0
3=6k
3/6 =k
1/2 =k
I hope this will help you
-by ABHAY
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