find the value of k such that quadratic polynomial 3x2+2kx+x-k-5 has sun of zeroes as half of their products
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Answered by
865
Hi ,
Let p( x ) = 3x² + 2kx + x - k - 5
= 3x² + ( 2k + 1 ) x - ( k + 5 )
compare p( x ) with ax² + bx + c
a = 3 , b = 2k + 1 , c = - ( k + 5 )
1 ) sum of the zeroes = - b / a
= - ( 2k + 1 ) / 3
2 ) product of the zeroes = c /a
= - ( k + 5 ) / 3
according to the problem given ,
- ( 2k + 1 ) / 3 = 1/2 [ - ( k + 5 ) / 3 ]
2 ( 2k + 1 ) = ( k + 5 )
4k + 2 = k + 5
4k - k = 5 - 2
3k = 3
k = 3/3
k = 1
I hope this helps you.
:)
Let p( x ) = 3x² + 2kx + x - k - 5
= 3x² + ( 2k + 1 ) x - ( k + 5 )
compare p( x ) with ax² + bx + c
a = 3 , b = 2k + 1 , c = - ( k + 5 )
1 ) sum of the zeroes = - b / a
= - ( 2k + 1 ) / 3
2 ) product of the zeroes = c /a
= - ( k + 5 ) / 3
according to the problem given ,
- ( 2k + 1 ) / 3 = 1/2 [ - ( k + 5 ) / 3 ]
2 ( 2k + 1 ) = ( k + 5 )
4k + 2 = k + 5
4k - k = 5 - 2
3k = 3
k = 3/3
k = 1
I hope this helps you.
:)
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161
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