Question

find the value of k such that the polynomial x^2-(k+6)x+2(2k-1) has sum of its zeros equal to half of their product​

Answer 1

Answer :-

x² - (k+6)x + 2(2k-1)

As sum of zeros

= -b/a

= -(-(k+6))/1

= k+6

Product of zeros

= c/a

= 2(2k-1)/1

= 2(2k-1)

According to the question

Sum of zeros =1/2 Product of zeros

=> -b/a = 1/2 × c/a

=> k + 6 = 1/2 × 2(2k -1)

=> k + 6 = 2k -1

=> 6 + 1 = 2k - k

=> 7 = k

So Value of k = 7

Answer 2

Answer:

Step-by-step explanation:

Sum of zero = (-b)/a

= -(-(k+6))

=k+6

Product of zero

=c/a

= 2(2k-1)

=4k-2

It is given in question

Sum of zero = 1/2 of product

K+6 = 1/2(4k-2

k+6 = 2k -1

6+1 = 2k-k

7 = k

Therefore k = 7