Question

if one of the zeros of the q.p (k-2)x2+(k-1)x-4 is -2then find the value of II 0​

Answer 1

Answer x=

$\frac{2}{3}$

Steps :

$(k - 2) {x}^{2} + (k - 1)x - 4 = 0 \\ \\ p( - 2) = (k - 2)4 +( k - 1)( - 2) - 4 = 0 \\ \\ 4k - 8 - 2k + 2 - 4 = 0 \\ 2k = 10 \\ k = 5 \\ using \: k = 5 in \: p(x) \\ 3 {x}^{2} + 4x - 4 = 0 \\ 3 {x}^{2} + 6x - 2 x - 4 = 0 \\ 3x(x + 2) - 2(x + 2) = 0 \\ (3x - 2)(x + 2) = 0 \\ x = - 2 \: \\ x = \frac{2}{3}$

hope this will help you....