plz solve this by giving whole solution
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hi here is ur awswer
ans=2
hope it helps
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ans=2
hope it helps
mark as brainliest
Answered by
2
Hey Mate !
Here is your solution :
Given,
p( x ) = x² - ( 2k - 2 )x + ( 2k + 1 )
Product of zeroes = ( 3/2 ) Sum of zeroes
Here,
Coefficient of x² ( a ) = 1
Coefficient of x ( b ) = -( 2k -2 )
Constant term ( c ) = ( 2k + 1 )
Let , α and β be its zeroes.
We know the relationship between zeroes and coefficients of x.
=> Sum of zeroes = -b/a
=> α + β = -[ - ( 2k - 2 ) ] / 1
=> α + β = ( 2k - 2 )
And,
=> Product of zeroes = c/a
=> αβ = ( 2k + 1 ) / 1
=> αβ = ( 2k + 1 )
Now,
=> Product of zeroes = ( 3/2 ) Sum of zeroes
=> αβ = ( 3/2 ) ( α + β )
By substituting the value of ( α + β ) and αβ.
=> ( 2k + 1 ) = ( 3/2 ) ( 2k - 2 )
=> ( 2k + 1 ) = ( 3/2 ) 2 ( k - 1 )
=> ( 2k + 1 ) = 3 ( k - 1 )
=> 2k + 1 = 3k - 3
=> 1 + 3 = 3k - 2k
=> 4 = k
Hence , the answer is option ( d ) 4.
=================================
Hope it helps !! ^_^
Here is your solution :
Given,
p( x ) = x² - ( 2k - 2 )x + ( 2k + 1 )
Product of zeroes = ( 3/2 ) Sum of zeroes
Here,
Coefficient of x² ( a ) = 1
Coefficient of x ( b ) = -( 2k -2 )
Constant term ( c ) = ( 2k + 1 )
Let , α and β be its zeroes.
We know the relationship between zeroes and coefficients of x.
=> Sum of zeroes = -b/a
=> α + β = -[ - ( 2k - 2 ) ] / 1
=> α + β = ( 2k - 2 )
And,
=> Product of zeroes = c/a
=> αβ = ( 2k + 1 ) / 1
=> αβ = ( 2k + 1 )
Now,
=> Product of zeroes = ( 3/2 ) Sum of zeroes
=> αβ = ( 3/2 ) ( α + β )
By substituting the value of ( α + β ) and αβ.
=> ( 2k + 1 ) = ( 3/2 ) ( 2k - 2 )
=> ( 2k + 1 ) = ( 3/2 ) 2 ( k - 1 )
=> ( 2k + 1 ) = 3 ( k - 1 )
=> 2k + 1 = 3k - 3
=> 1 + 3 = 3k - 2k
=> 4 = k
Hence , the answer is option ( d ) 4.
=================================
Hope it helps !! ^_^
Anonymous:
:-)
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