Question

5) a,ß and gamma are zeros of p(x)=kx2-5x+9 and a3+ß3+r3=27 , find k

Answer 1

Answer:

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Step-by-step explanation:

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