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Let f(x)=ax^(2)+bx+c, g(x)=ax^(2)+px+q, ...

Let `f(x)=ax^(2)+bx+c`, `g(x)=ax^(2)+px+q`, where `a`, `b`, `c`, `q`, `p in R` and `b ne p`. If their discriminants are equal and `f(x)=g(x)` has a root `alpha`, then

A

`alpha` will be `A.M.` of the roots of `f(x)=0`, `g(x)=0`

B

`alpha` will be `G.M.` of the roots of `f(x)=0`, `g(x)=0`

C

`alpha` will be `A.M.` of the roots of `f(x)=0` or `g(x)=0`

D

`alpha` will be `G.M.` of the roots of `f(x)=0` or `g(x)=0`

Text Solution

Verified by Experts

The correct Answer is:
A
`(a)` `a alpha^(2)+b alpha+c=a alpha^(2)+p alpha+q=0`
`implies alpha=(q-c)/(b-p)`…………`(i)`
and `b^(2)-4ac=p^(2)-4aq`
`impliesb^(2)-p^(2)=4a(c-q)`
`implies b+q=(4a(c-q))/(b-p)=-4a alpha` (from `(i)`)
`:.alpha=(-(b+p))/(4a)=((-b)/(a)-(p)/(a))/(4)` which is `AM` of all the roots of `f(x)=0` and `g(x)=0`
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