
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
161.1k+ views
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.
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.
Recently Updated Pages
If tan 1y tan 1x + tan 1left frac2x1 x2 right where x frac1sqrt 3 Then the value of y is

Magnetic Force on a Current-Carrying Wire - Important Concepts and Tips for JEE

A steel rail of length 5m and area of cross section class 11 physics JEE_Main

At which height is gravity zero class 11 physics JEE_Main

A nucleus of mass m + Delta m is at rest and decays class 11 physics JEE_MAIN

Two pi and half sigma bonds are present in A N2 + B class 11 chemistry JEE_Main

Trending doubts
JEE Main 2026 Syllabus PDF - Download Paper 1 and 2 Syllabus by NTA

JEE Main Eligibility Criteria 2025

JEE Main B.Arch Cut Off Percentile 2025

JoSAA Counselling 2025: Registration Dates OUT, Eligibility Criteria, Cutoffs

NIT Cutoff Percentile for 2025

JEE Mains 2025 Cutoff: Expected and Category-Wise Qualifying Marks for NITs, IIITs, and GFTIs

Other Pages
NCERT Solutions for Class 10 Maths Chapter 13 Statistics

NCERT Solutions for Class 10 Maths Chapter 11 Areas Related To Circles

NCERT Solutions for Class 10 Maths Chapter 12 Surface Area and Volume

NCERT Solutions for Class 10 Maths Chapter 14 Probability

List of Fastest Century in IPL History

NEET 2025 – Every New Update You Need to Know
