Question

Let $f\left(x\right)=a{x}^2+bx+c,g\left(x\right)=a{x}^2+px+q$ where $a,b,c,q,p,\in R\&b\ne p.$ If their discriminants are equal and $f\left(x\right)=g\left(x\right)$ has a root $\alpha$ then

• A. None of the above
• 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 $f\left(x\right)=0andg\left(x\right)=0$
• D. $\alpha$ will be $A.M$ of the roots of $g\left(x\right)=0$

Answer 1

The correct option is C

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

$a{\alpha }^2+b\alpha +c=a{\alpha }^2+p\alpha +q\Rightarrow \alpha =\frac{q-c}{b-p}\to \left(i\right)And{b}^2-4ac=p^2-4aq\Rightarrow b^2-p^2=4a(c-q)\Rightarrow b+p=\frac{4a(c-q)}{b-p}=-4a\alpha \left(from\right(i\left)\right)$

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

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

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

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

Now, $A.Mof{x}_1,x_2,x_3,x_{4}is\frac{x_1+x_2+x_3+x_4}{4}$

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

Hence, $\alpha =$ will be $A.M.$ of the roots of $f\left(x\right)=0andg\left(x\right)=0$