Question

If the roots of the equation $a{x^2} + bx + c = 0$ be $\alpha $ and $\beta $ then the roots of the equation $c{x^2} + bx + a = 0$ are:
A. $ - \alpha , - \beta $
B. $\alpha ,\dfrac{1}{\beta }$
C. $\dfrac{1}{\alpha },\dfrac{1}{\beta }$
D. None of these

Answer 1

Hint: First we will find the sum and product of roots of equation $a{x^2} + bx + c = 0$. Then divide sum by product of roots. After simplification, we will compare the sum and product of the roots of equation $c{x^2} + bx + a = 0$ to find the roots of the equation.

Formula Used: If the quadratic equation is $a{x^2} + bx + c = 0$. Then,
Sum of the roots $ = \dfrac{{ - b}}{a}$
Product of roots $ = \dfrac{c}{a}$

Complete step by step solution: Given, the roots of the equation $a{x^2} + bx + c = 0$ are $\alpha $ and $\beta $.
Sum of the roots $ = \dfrac{{ - b}}{a}$
$\alpha + \beta = \dfrac{{ - b}}{a}$ …(1)
Product of roots $ = \dfrac{c}{a}$
\[\alpha \beta = \dfrac{c}{a}\] …(2)
Dividing equation (1) by equation (2)
$\dfrac{{\alpha + \beta }}{{\alpha \beta }} = \dfrac{{\dfrac{{ - b}}{a}}}{{\dfrac{c}{a}}}$
$\dfrac{1}{\alpha } + \dfrac{1}{\beta } = \dfrac{{ - b}}{c}$ …(3)
Taking reciprocal of equation (2)
$\dfrac{1}{{\alpha \beta }} = \dfrac{a}{c}$ …(4)
Given equation, $c{x^2} + bx + a = 0$
Sum of roots op equation $c{x^2} + bx + a = 0$ is $\dfrac{{ - b}}{c}$
Product of roots op equation $c{x^2} + bx + a = 0$ is $\dfrac{a}{c}$
From equation (3) and (4)
Roots of the equation $c{x^2} + bx + a = 0$ are $\dfrac{1}{\alpha },\dfrac{1}{\beta }$

Hence, correct option is (C)

Note: Students should divide sum by product of roots of equation for easy calculations if they try any other method they can make mistakes.