Consider two quadratic expressions `f(x) =ax^2+ bx + c and g (x)=ax^2+px+c,( a, b, c, p,q in R, b != p)` such that their discriminants are equal. If `f(x)= g(x)` has a root `x = alpha`, then

Options

A. `alpha` will be AM of the roots of `f(x)=0` and `g(x)=0`
B. `alpha` will be AM of the roots of `f(x)=0`
C. `alpha` will be AM of the roots of `f(x)=0` or `g(x)=0`
D. `alpha` will be AM of the roots of `g(x)=0`

Answer

A

Explanation

Given `b^(2)-4ac=p^(2)-4aq`……….i
and `f(x)=g(x)`
`impliesax^(2)+bx+c=ax^(2)+px+q`
`implies(p-p)x=q-c`
`:.x=(q-c)/(b-p)=alpha` [given (ii)]
From Eq. (i) we get
`(b+p-)(b-p)+4a(q-c)=0`
`implies(p+p)(b-p)+4a alpha(b-p)=0` [from Eq. (ii)]
or `alpha=-((p+p))/(4a)[ :' b!=p]`
`=((-b/a)+(-p/a))/4` ltrbgt `=(["Sum of theroots of" (f(x)=0)+ "Sum of the roots of"(g(x)=0)])/4`
`=AM` of the roots of `f(x)=0`
and `g(x)=0`