If a,b,c are in G.P.,then the equations $a{x}^2+2bx+c=0$ and $d{x}^2+2ex+f=0$ have a common root if ${\frac{d}{a}}^{\prime }{\frac{e}{b}}^{\prime }\frac{f}{c}$ are in

(i) A. P.

(ii) G.P.

(iii) H.P.

(iv) None of these.

Another form :

If $a{x}^2+2bx+c=0$ and $p{x}^2+2qx+r=0$ have a common root and ${\frac{a}{p}}^{\prime }{\frac{b}{q}}^{\prime }\frac{c}{r}$ in A.P. then prove that p,q,r are in G.P.

(i).

a,b,c in G.P. $\Rightarrow b^2-ac=0$

${\Delta }_1=4(b^2-ac)=0$ $\therefore$ It has equal roots

$\therefore$ $\alpha +\alpha =-\frac{2b}{a}$ or $\alpha =-\frac{b}{a}$

Since the two equations have common roots therefore

$\alpha =-b/a$ must also be a root of 2nd equation.

$\therefore$ $d{(-\frac{b}{a})}^2+2e(-\frac{b}{a})+f=0$

or $\frac{d}{a}.\frac{b^2}{a}+\frac{2e}{b}\left(\frac{{-b}^2}{a}\right)+\frac{f}{c}c=0$ (Note)

or $\frac{d}{a}.\frac{b^2}{a}-\frac{2e}{b}\left(\frac{b^2}{a}\right)+\frac{f}{c}\left(\frac{b^2}{a}\right)=0$

$\because$ $b^2=ac$

or $\frac{d}{a}-\frac{2e}{b}+\frac{f}{c}=0$ or $\frac{d}{a},\frac{e}{b},\frac{f}{c}$ are in A.P.

Another form : Assume that p,q,r are in G.P., so that $q^2=pr$ which implies that roots of 2nd equation are equal and rest as above.