Question

If alpha,beta,gamma are the zeroes of kx^3-5x+9 and alpha^3+beta^3+gamma^3=27,find 'k'

Answer 1

By using the relation between zeroes and coefficients, we have:

alpha + beta + gamma = -0/k = 0

alpha . beta . gamma = -9/k

 

Therefore,

alpha3 + beta3 + gamma3 = 3alpha . beta . gamma

So, 27 = 3 x -9/k

27 = -27/k

k = -1