Question

If the sum of the zeroes of the polynomial x² -(k+6)x +2(2k-1) is half of their product, find tu value if k

Answer 1

Hey

Here is your answer,

x² -(k+6)x +2(2k-1) = 0

Sum of zeroes = -b/a
= k+6/1
= k+6

Product of zeroes = c/a
= 2(2k-1)/1
= 2(2k-1)

Given that,

Sum of zeroes = 1/2 Product of zeroes
k+6 = 2(2k-1)/2
2k+12 = 4k-2
2k-4k = -2-12
-2k = -14
k = 14/2
k = 7

Hope it helps you!

Answer 2

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




And,


Product of zeroes = C/A

Alpha × Beta = 4K - 2 /1




Alpha × Beta = 4K - 2 .




According to the question,


Sum of zeroes = 1/2 × Product of zeroes



K + 6 = 1/2 × (4K - 2 )








2 ( K + 6 ) = 4K - 2








2K + 12 = 4K - 2



2K = 14



K = 14/2



K = 7