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)`.

Explanation

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