Question

if the zeroes of quadratuc polynomial p(x) = 3x^2 + (2k-1)x - 5 are equal in magnitude but opposite in sign then find the value of k.

Answer 1

let zeros be a and-a
put x=a and x=-a
then we get
3a^2 +(2k-1)a-5=3 (-a )^2+(2k-1)-a-5
3a^2+(2k-1)a=3a^2-(2k-1)a
2k-1=-2k+1
4k=2
k=1/2