Question

The roots of $a_1{x}^2+b_{1}x+c_2=0$ are reciprocal of the roots of the equation $a_2{x}^2+b_{2}x+c_2=0$

• A. $\frac{a_1}{a_2}=\frac{b_1}{b_2}=\frac{c_1}{c_2}$
• B. $\frac{b_1}{b_2}=\frac{c_1}{a_2}=\frac{a_1}{c_2}$
• C. $\frac{a_1}{a_2}=\frac{b_1}{c_2}=\frac{c_1}{b_2}$
• D. $a_1=\frac{1}{a_2},b_1=\frac{1}{b_2},c_1=\frac{1}{c_2}$

Answer 1

The correct option is B $\frac{b_1}{b_2}=\frac{c_1}{a_2}=\frac{a_1}{c_2}$

Given:$a_1{x}^2+b_{1}x+c_1=0$ ........$\left(1\right)$

$a_2{x}^2+b_{2}x+c_2=0$ ........$\left(2\right)$

Let $\alpha ,\beta$ be the roots of $a_1{x}^2+b_{1}x+c_1=0$ ........$\left(1\right)$

$\Rightarrow \alpha +\beta =-\frac{b_1}{a_1}$ and

$\alpha \beta =\frac{c_1}{a_1}$

Given:Roots of $\left(1\right)$ are reciprocal to $\left(2\right)$

$\frac{1}{\alpha }+\frac{1}{\beta }=-\frac{b_2}{a_2}$ and $\frac{1}{\alpha \beta }=\frac{c_2}{a_2}$

$\Rightarrow \frac{\alpha +\beta }{\alpha \beta }-\frac{b_2}{a_2}$ and $\frac{1}{\alpha \beta }=\frac{c_2}{a_2}$

Using $\alpha +\beta =-\frac{b_1}{a_1}$ and $\alpha \beta =\frac{c_1}{a_1}$ we have

$\Rightarrow \frac{-\frac{b_1}{a_1}}{\frac{c_1}{a_1}}=-\frac{b_2}{a_2}$ and $\frac{1}{\frac{c_1}{a_1}}=\frac{c_2}{a_2}$

$\Rightarrow \frac{-b_1}{c_1}=-\frac{b_2}{a_2}$ and

$\frac{a_1}{c_1}=\frac{c_2}{a_2}$

$\Rightarrow \frac{b_1}{b_2}=\frac{c_1}{a_2}$ and

$\frac{a_1}{c_1}=\frac{c_2}{a_2}$

$\Rightarrow \frac{b_1}{b_2}=\frac{c_1}{a_2}$ and

$\frac{c_1}{a_1}=\frac{a_2}{c_2}$

$\Rightarrow \frac{b_1}{b_2}=\frac{c_1}{a_2}$ and

$\frac{c_1}{a_2}=\frac{a_1}{c_2}$

$\therefore \frac{b_1}{b_2}=\frac{c_1}{a_2}=\frac{a_1}{c_2}$

Option$\left(b\right)$ is correct.