the zeros of quadratic polynomial X square + 4 x + k are alpha and beta evaluate the values of k if alpha plus 2 beta -1 About the author Amara
[tex]\large{\pmb{\underline{\underline{\sf{Answer~-}}}}}[/tex] [tex]\huge{\underline{\underline{\frak{\pink{—}}}}}[/tex] [tex]\dashrightarrow[/tex] given that α+β = αβ where α and β are the roots of the equation x² – (k-6)x +(2k +1) = 0 solution:- we know that; α+β = – (coefficient of x )/( coefficient of x² ) = and αβ = (constant term)/(coefficient of x²) = in the question it is given that α+β = αβ => – (coefficient of x )/( coefficient of x² ) =(constant term)/(coefficient of x²) => -{-(k -6) } / 1 = (2k +1) / 1 => (k – 6) = 2k +1 => 2k – k = -6 -1 => k = – 7answer Reply
[tex]\large{\pmb{\underline{\underline{\sf{Answer~-}}}}}[/tex]
[tex]\huge{\underline{\underline{\frak{\pink{—}}}}}[/tex]
[tex]\dashrightarrow[/tex]
given that
α+β = αβ
where α and β are the roots of the equation x² – (k-6)x +(2k +1) = 0
solution:-
we know that;
α+β = – (coefficient of x )/( coefficient of x² )
=
and αβ = (constant term)/(coefficient of x²)
=
in the question it is given that
α+β = αβ
=> – (coefficient of x )/( coefficient of x² ) =(constant term)/(coefficient of x²)
=> -{-(k -6) } / 1 = (2k +1) / 1
=> (k – 6) = 2k +1
=> 2k – k = -6 -1
=> k = – 7answer
Step-by-step explanation:
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