Question

The value of k such that $3x^2+2kx+x-k-5$ have the sum of the zeroes as half of their product,

• A. 4
• B. 3
• C. 2
• D. 1

Answer 1

The correct option is

D

1

Sum of zeros = $\frac{-(2k+1)}{3}$

Product of zeros = $\frac{-(k+5)}{3}$

Given that,$\frac{-(2k+1)}{3}$= -$\frac{1}{2}$ $\frac{(k+5)}{3}$

2(2k + 1) = k + 5

4k + 2 = k + 5

3k = 3 ,

Therefore, k = 1