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 I need a well explained solution for a and b
Thanx soo much

Answer 1

Given equation is x^2-px+8=0

a. alpha-beta= root under (b^2-4ac)÷a

so 2=root under (p^2-32)÷1

taking square in both side

4=p^2-32

p^2=36

so p= 6

b.alpha^3+beta^3=(alpha +beta)(alpha^2-alpha.beta+beta^2)

since alpha +beta=-b/a=-(-6)÷1=6

alpha.beta=c/a=8÷1=8

same will find alpha^2+beta^2=(alph a+beta)^2-2.alpha.beta

and last we will put all these value in equation alpha^3+beta^3 and will get answer