Let p and q real number such that `p!= 0`,`p^3!=q` and `p^3!=-q`. if `alpha` and `beta` are non-zero complex number satisfying `alpha+beta=-p` and `alpha^3+beta^3=q`, then a quadratic equation having `alpha/beta` and `beta/alpha` as its roots is

Options

A. `(p^(3)+q)x^(2)-(p^(3)+2q)x+(p^(3)+q)=0`
B. `(p^(3)+q)x^(2)-(p^(3)-2q)x+(p^(3)+q)=0`
C. `(p^(3)-q)x^(2)-(5p^(3)-2q)x+(p^(3)-q)=0`
D. `(p^(3)-q)x^(2)-(5p^(3)+2q)x+(p^(3)-q)=0`

Answer

B

Explanation

We have, `alpha+beta = -p and alpha^(3) + beta^(3) = q`
`therefore" "(alpha+beta)^(3) = alpha^(3)+beta^(3) + 3 alpha beta (alpha + beta)`
`rArr" "-p^(3) = q - 3 alpha beta p`
`rArr" "alpha beta = (p^(3)+q)/(3p)`
`(alpha)/(beta)+(beta)/(alpha)=(alpha^(2)+beta^(2))/(alpha beta)=((alpha+beta)^(2))/(alpha beta)-2=(3p^(3))/(p^(3)+q)-2=(p^(3)-2p)/(p^(3)+q)`
The quadratic equation having `(alpha)/(beta) and (beta)/(alpha)` as its roots is `x^(2)-((alpha)/(beta)+(beta)/(alpha))x + (alpha)/(beta)xx(beta)/(alpha)=0`
`rArr" "x^(2)-((p^(3)-2p)/(p^(3)+q))x+1 = 0`
`rArr" "(p^(3) +q) x^(2) -(p^(3)-2q) x + (p^(3)+q)=0`