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Consider quadratic equations x^2-ax+b=0 ...

Consider quadratic equations `x^2-ax+b=0 and x^2+px+q=0` If the above equations have one common root and the other roots are reciprocals of each other, then `(q-b)^2` equals

A

bq(p-a)^(2)`

B

`b(p-a)^(2)`

C

`q(p-a)^(2)`

D

none of these

Text Solution

Verified by Experts

The correct Answer is:
A
`(a)` x^(2)-ax+b=0`……..`(i)`
x^(2)-px+q=0`……….`(ii)`
Let the roots of `(i)` be `alpha`, `beta` and that of `(ii)` be `alpha`, `(1)/(beta)`
`:. Alpha+beta=a`, `alphabeta=b`, `alpha+(1)/(beta)=p`, `(alpha)/(beta)=q`
`:.((q-b)^(2))/((p-a)^(2))=(alpha^(2)((1)/(beta)-beta)^(2))/(((1)/(beta)-beta)^(2))`
But `alpha^(2)=alphabeta*(alpha)/(beta)=bq`
`implies (q-b)^(2)=bq(p-a)^(2)`
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