Question

Find the value of k such that the polynomial x²-(k+6) X+2(2k-1) has sum of its zeroes equal to half of their product​

Answer 1

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

Answer 2

$\bf\huge\ Question:-$

Find the value of k such that the polynomial x²-(k+6) X+2(2k-1) has sum of its zeroes equal to half of their product

$\bf\huge\ Answer:-$

Let the roots be α,β

For a quadratic equation ax² +bx+c=0

Sum of roots α+β=k+6⋯(1)

Product of roots αβ=4k−2⋯(2)

According to question,

α+β= αβ/2

From (1),(2)

⟹k+6=2k−1

⟹k=7