InterviewSolution
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InterviewSolution
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If `alpha,beta` are roots of `x^2-px+q=0` then find the quadratic equation whose roots are `((alpha^2-beta^2)(alpha^3-beta^3))` and `alpha^2beta^3+alpha^3beta^2` |
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Answer» Since `alpha, beta` are the roots of `x^(2)-px+q=0` `:. alpha+beta=p,alphabeta=q` `impliesalpha-beta=sqrt((p^(2)-4q))` Now `(alpha^(2)-beta^(2))(alpha^(3)-beta^(3))` `=((alpha+beta)(alpha-beta)(alpha-beta)(alpha^(2)+alpha beta+ beta^(2))` `=(alpha+beta)(alpha-beta)^(2){(alpha+beta)^(2)-alpha beta}` `=p(p^(2)-4q)(p^(2)-q)` and `alpha^(3)beta^(2)+alpha^(2)beta^(3)=alpha^(2)beta^(2)(alpha+beta)=pq^(2)` `S=` Sum of roots `=p(p^(2)-4q)(p^(2)-q)+pq^(2)` `=p(p^(4)-5p^(2)q+5q^(2))` `P=` Product of roots `=p^(2)q^(2)(p^(2)-4q)(p^(2)-q)` `:.` Required equation is `x^(2)-Sx+P=0` i.e. `x^(2)-p(p^(4)-5p^(2)q+5q^(2))x+p^(2)q^(2)(p^(2)-4q)(p^(2)-q)=0` |
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