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Q. Let p and q real number such that p!=...

Q. Let p and q real number such that `p!= 0`,`p^2!=q` and `p^2!=-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

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`

Text Solution

Verified by Experts

The correct Answer is:
B
Sum of roots `=(alpha^(2)+beta^(2))/(alphabeta)` and product = 1
Given, `alpha+beta=-pand alpha^(3)+beta^(3)=q`
`implies(alpha+beta)(alpha^(2)-alphabeta+beta^(2))=q`
`thereforealpha^(2)+beta^(2)-alphabeta=(-q)/(p)" "....(i)`
and `(alpha+beta)^(2)=p^(2)`
`impliesalpha^(2)+beta^(2)+2alphabeta=p^(2)" "(ii)`
From Eqs. (i) and (ii) we get
`alpha^(2)+beta^(2)=(p^(a)-2q)/(3p)and alpha beta=(p^(3)+q)/(3p)`
`therefore` Required equation is, `x^(2)((p^(3)-2q)x)/((p^(3)+q))+1=0`
`implies(p^(3)+q)x^(2)-(p^(3)-2q)x+(p^(3)+q)=0`
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