Question

If $b^2=ac,$equation $a{x}^2+2bx+c=0$ and $d{x}^2+2ex+f=0$

have common roots then $\left(\frac{d}{a}\right),\left(\frac{e}{b}\right),\left(\frac{f}{c}\right)$ are in

• A. $A.P.$
• B. $G.P.$
• C. $H.P$
• D. $Noneofthese$

Answer 1

The correct option is A

$A.P.$

Explanation for the correct option:

Given,

$a{x}^2+2bx+c=0$ and

$b^2=ac$

$x=\frac{[-2\mathrm{b}\pm \sqrt{4{\mathrm{b}}^2-4\mathrm{ac}}]}{2a}$

$=-\frac{b}{a}$

Now $x=-\frac{b}{a}$ is also roots of

$d{x}^2+2ex+f=0$

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

$\Rightarrow$$\left(\frac{2\mathrm{eb}}{a}\right)=\left(\frac{{\mathrm{db}}^2}{a^2}\right)+f$

$\Rightarrow$ $2eb=dc+af$

$\Rightarrow$ $\left(\frac{2\mathrm{e}}{b}\right)=\left(\frac{\mathrm{d}}{\mathrm{a}}\right)+\left(\frac{\mathrm{f}}{\mathrm{c}}\right)$

$\Rightarrow$$\left(\frac{\mathbf{d}}{\mathit{a}}\right)\mathbf{,}\left(\frac{e}{b}\right)\mathbf{,}\left(\frac{f}{c}\right)$ are in AP

Hence option(A) is correct.