Question

Find the value of k such that the quadratic polynomial x^2-(k+6)x+2(2k+1) has sum of the zeroes is half of their product

Answer 1

Heya !!!




P(X) => X²-(K+6)X + 2(2K+1)



Here,



A => 1 , B = -K - 6 and C = 2(2K+1)





Sum of zeroes => -B/A




Alpha + Beta = -( - K-6)/ 1




Alpha + Beta = K + 6 ------- (1)




And,




Product of zeroes = C/A




Alpha × Beta =>2(2K+1)/1




Alpha × Beta => 4K +2 ---------(2)





According to question,




Sum of zeroes => 1/2 × ( Product of zeroes )



K +6 = 1/2 × ( 4K +2)




K +6 = 4K + 2/2




2( K +6) = 4K +2




2K + 12 => 4K +2





4K -2K => 12-2




2K = 10




K => 10/2



K => 5




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