Question

if alfa and beta are the zeros of the polynomial, x²-px+q. prove that alfa²/beta² + beta ²/alfa² = p²/q² - 4p²/q² +2

Answer 1

Given polynomial, x²-px+q
COMPARING this quadratic equation with standard quadratic polynomial ax^2+bx+c, we get a=1, b= -p, c=q
Since alfa and beta are the zeros of the polynomial,
alpha+beta= -(b/a)= -(-p/1)=p  alpha*beta=c/a =q/1= q
= (alpha^2/beta^2) + (beta^2/alpha^2) = (alpha^4+beta^4)/(alpha^2*beta^2)
= (alpha^2+beta^2)^2 - 2alpha^2*beta^2/(alpha^2*beta^2)
= [(alpha+beta)^2 - 2alpha*beta]^2 - 2(alpha*beta)^2/(alpha*beta)^2
= [p^2- 2q]^2 - 2q^2/q^2
= (p^4+ 4q^4 - 4p^2q -2q^2)/q^2
= (p^4 - 4p^2q +2q^2)/q^2
= p^4/q^2 - 4p^2q/q^2 + 2q^2/q^2
= p^4/q^2 - 4p^2/q + 2