Question

The roots of the equation x^2-px+8=0 are alpha and beta. If the roots differ by 2
a. Calculate the possible values of p
b. Hence, find the possible values of alpha^3+beta^3

Pls. It's really urgent. A step by step and well explained solution pls

Answer 1

Answer:

p = 6 & -6

alpha^3 + Beta^3 =  72   &  -72  respectively

Step-by-step explanation:

x^2 - px + 8 = 0

roots are alpha & Beta

alpha = Beta + 2

(x - alpha)(x-Beta) = 0

(x - (Beta+2))(x-Beta) = 0

x^2 - x Beta - x Beta - 2x  + Beta^2 + 2Beta

x^2 - x (2beta + 2) + (beta^2 + 2Beta)

so p = 2beta + 2

beta^2 +2Beta  = 8

Beta^2 + 4beta - 2Beta - 8 = 0

(Beta+4)(Beta-2) = 0

Beta = -4  & 2

alpha = Beta +2 =  -2 , 4  (respectively for beta = -4 & 2 )

(alpha , Beta )  = (-2 , -4)  & (4 ,2)

p = 2beta + 2 = -6 for Beta = -4

= 6 for Beta = 2

alpha^3 + Beta ^3

(-2)^3 + (-4)^3 = -8 - 64 = -72

4^3 + 2^3 = 64 + 8 = 72

Answer 2

Answer:

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