InterviewSolution
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InterviewSolution
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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 isA. `(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` |
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Answer» Correct Answer - C `(alpha)/(beta)+(beta)/(alpha)=(alpha^(2)+beta^(2))/(alpha beta)=((alpha +beta)^(2)-2 alpha beta)/(alpha beta)`……….i and given `alpha^(3)+beta^(3)=q, alpha + beta=-p` `implies(alpha+beta^(3)=q,alpha + beta=-p` `=(alpha +beta)^(3)-3alpha beta(alpha +beta)=q` `implies-p^(3)+3p alpha beta=q` ltbr or `alpha beta=(q+p^(3))/(3p)` `:.` From eq. (i) we get `(alpha)/(beta)+(beta)/(alpha)=(p^(2)-(2(q+p^(3)))/(3p)/(((q+p^(3)))/(3p))=(p^(3)-2q)/((q+p^(3)))` and product of the roots `=(alpha)/(beta). (beta)/(alpha)=1` `:.` Required equation is `x^(2)-((p^(3)-2q)/(q+p^(3)))x+q=0` or `(q+p^(3))x^(2)-(p^(3)-2q)x+(q+p^(3))=0` |
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