Question

if the sum of zeros of 8 x square -kx + 1 is -1 then find the value of k​

Answer 1

$\alpha + \beta = - 1$

$p(x) = 8 {x}^{2} - kx + 1$

a= 8; b= -k ; c= 1

$\alpha + \beta = \frac{ - b}{a}$

$- 1 = \frac{ - k}{8}$

$- k \: = ( - 1) \times (8)$

-k = -8

Therefore ,. k=8