InterviewSolution
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InterviewSolution
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If the equation `x^(2)-px+q=0` and `x^(2)-ax+b=0` have a comon root and the other root of the second equation is the reciprocal of the other root of the first, then prove that `(q-b)^(2)=bq(p-a)^(2)`. |
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Answer» Let `alpha` and `beta` be the roots of `x^(2)-px+q=0`.Then `alpha+beta=p` …….i `alpha beta=q` ……………..ii And `alpha` and `1/(beta)` be the roots of `x^(2)-ax+b=0`. Then `alpha+1/(beta)=a` ……….iii `(alpha)/(beta)=b` ……..iv Now LHS`=(q-b)^(2)` `=(apha beta-(alpha)/(beta))^(2)` [from Eqs (ii) and(iv) ] `=alpha^(2)(beta-1/(beta))^(2)=alpha^(2)[(alpha+beta)-(alpha+1/(beta))]^(2)` `=alpha^(2)(p-a)^(2)` [from Eqs (i) and (iii)] `=apha .beta . (alpha)/(beta)(p-a)^(2)` `=pq(p-a)^(2)` [from Eqs (ii) and (iv) ] `=`RHS |
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