Question

If $a,b$ and $c$ are in geometrical progression and the roots of the equations $a{x}^2+2bx+c=0$ are $\alpha$ & $\beta$ and those of $e{x}^2+2bx+a=0$ are $\gamma$ and $\delta$, then

• A. $\alpha \ne \beta \ne \gamma \ne \delta$
• B. $\alpha \ne \beta and\gamma \ne \delta$
• C. $a\alpha =a\beta =c\gamma =c\delta$
• D. $\alpha =\beta and\gamma =\delta$

Answer 1

The correct option is A $\alpha \ne \beta \ne \gamma \ne \delta$

Given that $a,b$ and $c$ are in G.P.

let $r$ be the common ration then,

$a=a;b=ar;c=a{r}^2$

Also,

$\begin{array}{c}a{x}^2+2bx+c=0\\ \alpha +\beta =-\frac{2b}{a}=-\frac{2ar}{a}\Rightarrow -2r;\alpha \beta =\frac{c}{a}=\frac{a{r}^2}{a}\Rightarrow r^2\end{array}$

Then in,

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

$\gamma +\delta =\frac{-2b}{c}=\frac{-2ar}{a{r}^2}\Rightarrow \frac{-2}{r};\delta \gamma =\frac{a}{c}=\frac{ar}{a{r}^3}\Rightarrow \frac{1}{r^2}$

To find $\alpha$ and $\beta$

$\begin{array}{cc}(\alpha -\beta {)}^2& =(\alpha +\beta {)}^2-4\alpha \beta \\ & =4r^2-4r^2\\ & =0\\ \Rightarrow \alpha & =\beta \end{array}$

To find $\gamma$ and $\delta$

$\begin{array}{cc}(\gamma -\delta {)}^2& =(\gamma +\delta {)}^2-4\delta \gamma \\ & =\left(\frac{4}{r^2}\right)-\frac{4}{r^2}=0\\ \Rightarrow \gamma & =\delta \end{array}$

Now we know that,

$\begin{array}{cc}\alpha +\beta & =-2r\\ 2\alpha & =-2r\\ \Rightarrow \alpha & =-r,\beta =-\mu \end{array}$

Also we know that,

$r+\delta =-\frac{2}{r}$

$\begin{array}{cc}2\cdot f& =-\frac{2}{r}\\ \Rightarrow \delta & =\frac{-1}{4},\gamma =-\frac{1}{r}\end{array}$

Hence we conclude that,

$a\alpha =a\beta$ and $c\gamma =c\delta$

(C) is the correct answer.