Question

Consider two quadratic expressions $f\left(x\right)=a{x}^2+bx+c$ and $g\left(x\right)=a{x}^2+px+q$, $(b\ne q)$ such that their discriminants are equal. If $f\left(x\right)=g\left(x\right)$ has a root $x=\alpha$, then?

• A. $\alpha$ will be A.M. of the roots of $f\left(x\right)=0$ and $g\left(x\right)=0$
• B. $\alpha$ will be A.M. of the roots of $f\left(x\right)=0$
• C. $\alpha$ will be A.M. of the roots of $g\left(x\right)=0$
• D. None of these

Answer 1

Answer: (A)

$\alpha$ will be A.M. of the roots of $f\left(x\right)=0$ and $g\left(x\right)=0$

Discriminants equal means

$b^2-4ac=p^2-4aq$

Also given $x=\alpha$ for $f\left(x\right)=g\left(x\right)$

$b\alpha +c=p\alpha +q$

$\alpha (b-p)=q-c$

$\alpha =\frac{q-c}{b-p}$

As we have,

$b^2-p^2=4a(c-q)$

$b+p=-4a\alpha$

$\alpha =-\frac{b+p}{4a}$

Let the roots of $f\left(x\right)$ be $x_1,x_2$ and the roots of $g\left(x\right)$ be $x_3,x_4$

$x_1+x_2=-\frac{b}{a}$

$x_3+x_4=-\frac{p}{a}$

AM of $x_1,x_2,x_3,x_4$ is $\frac{x_1+x_2+x_3+x_4}{4}$

$\frac{x_1+x_2+x_3+x_4}{4}=-\frac{b+p}{4a}=\alpha$