Question

If $a,b,$ and $c$ are in geometric progression and the roots of the equation $a{x}^2+2bx+c=0$ are $\alpha ,\beta$ and those of $c{x}^2+2bx+a=0$ are $\gamma ,\delta ,$ 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 $\gamma \ne \delta$
• E. $\alpha \ne \beta$ and $\gamma –\delta$

Answer 1

The correct option is C

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

Explanation for the correct option:

Step 1. Find the roots of equation $\mathit{a}{\mathit{x}}^{\mathbf{2}}\mathbf{}\mathbf{+}\mathbf{}\mathbf{2}\mathit{b}\mathit{x}\mathbf{}\mathbf{+}\mathbf{}\mathit{c}\mathbf{}\mathbf{=}\mathbf{}\mathbf{0}$ :

Given that, $a,b,$ and $c$ are in geometric progression

$\Rightarrow$$a=a,b=ar,c=a{r}^2$, $r=$Common ratio

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

$\begin{array}{rcl}\alpha +\beta & =& \frac{–2b}{a}\\ & =& \frac{–2ar}{a}\\ & =& –2r\end{array}$

$\begin{array}{rcl}\alpha \beta & =& \frac{c}{a}\\ & =& \frac{a{r}^2}{a}\\ & =& r^2\end{array}$

Step 2. Find the roots of equation $c{x}^2+2bx+a=0$:

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

$\begin{array}{rcl}\gamma +\delta & =& \frac{–2b}{c}\\ & =& \frac{–2ar}{a{r}^2}\\ & =& \frac{–2}{r}\end{array}$

$\begin{array}{rcl}\gamma \delta & =& \frac{a}{c}\\ & =& \frac{a}{a{r}^2}\\ & =& \frac{1}{r^2}\end{array}$

Step 3. Find the value of $\alpha$ and $\beta$:

$\begin{array}{rcl}{(\alpha –\beta )}^2& =& {(\alpha +\beta )}^2–4\alpha \beta \\ & =& {\left(-2r\right)}^2-4r^2\\ & =& 4r^2-4r^2\\ & =& 0\end{array}$

$\Rightarrow$ $\alpha =\beta$

Step 4. Find the value of $\gamma$ and $\delta$:

$\begin{array}{rcl}{(\gamma –\delta )}^2& =& {(\gamma +\delta )}^2–4\gamma \delta \\ & =& {\left(\frac{-2}{r}\right)}^2-4\left(\frac{1}{r^2}\right)\\ & =& \frac{4}{r}-\frac{4}{r}\\ & =& 0\end{array}$

$\Rightarrow$ $\gamma =\delta$

Now, we know that

$\alpha +\beta =-2r$

$\Rightarrow$$2\alpha =-2r$

$\Rightarrow$ $\alpha =-r$

$\therefore$$a\alpha =–ar,a\beta =–ar$

Also we know that,

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

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

$\Rightarrow$ $\delta =\frac{-1}{r}$

$\therefore$$c\gamma =\frac{–a{r}^2}{r}=–ar,c\delta =–ar$

$\therefore$$\mathit{a}\mathit{\alpha }\mathbf{}\mathbf{=}\mathbf{}\mathit{a}\mathit{\beta }\mathbf{}\mathbf{=}\mathbf{}\mathit{c}\mathit{\gamma }\mathbf{}\mathbf{=}\mathbf{}\mathit{c}\mathit{\delta }$

Hence, Option ‘C’ is Correct.