find the value of k, (3k+1)(3k-2)
minnie147:
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Answered by
4
=9k^2-6k+3k-2=0
=9k^2-3k-2=0
=9k^2-6k+3k-2=0
=3k(3k-2)+1(3k-2)
=(3k-2)(3k+1)=0
k=2/3
ork=-1/3
k have two solutions k=2/3,-1/3;not k=3/2,-1/3
Answered by
3
Q] find the value of k, (3k+1)(3k-2)
ANS]
↪[ 3k + 1 ] [ 3k - 2 ]
= 9k^2 -6k + 3k - 2
= 9k^2 -3k - 2
= 9k ^2 - 6k + 3k -2
=3k [ 3k - 2 ] + 1 [ 3k - 2 ]
= [3k + 1 ] [ 3k - 2 ]
= k = -1/3. OR. k = 2/3
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