
If the roots of the equation $\dfrac{\alpha }{{x - \alpha }} + \dfrac{\beta }{{x - \beta }} = 1$ are equal in magnitude but opposite in sign, then $\alpha + \beta $ is equal to:
A. 0
B. 1
C. 2
D. 3
Answer
162.3k+ views
Hint: Expand the equation and write it in the standard form of a quadratic equation. Use the fact the sum of roots of a quadratic equation of the form $a{x^2} + bx + c = 0$ is $\dfrac{{ - b}}{a}$.
Complete step by step Solution:
Expand the equation $\dfrac{\alpha }{{x - \alpha }} + \dfrac{\beta }{{x - \beta }} = 1$
$\dfrac{{\alpha (x - \beta ) + \beta (x - \alpha )}}{{(x - \alpha )(x - \beta )}} = 1$
$\alpha (x - \beta ) + \beta (x - \alpha ) = (x - \alpha )(x - \beta )$
Expanding further,
$\alpha x - \alpha \beta + \beta x - \alpha \beta = {x^2} - \alpha x - \beta x + \alpha \beta $
${x^2} - 2\alpha x - 2\beta x + 3\alpha \beta = 0$
${x^2} - 2(\alpha + \beta )x + 3\alpha \beta = 0$
The above equation is of the form $a{x^2} + bx + c = 0$. We know that the sum of roots of a quadratic equation of the form $a{x^2} + bx + c = 0$ is $\dfrac{{ - b}}{a}$. It is given to us that the roots are equal in magnitude and opposite in sign. Therefore, the sum of roots must equal zero.
Therefore,
\[\;2(\alpha + \beta ) = 0\]
\[\;\alpha + \beta = 0\]
Therefore, the correct option is (A).
Note:Given an equation in the standard form $a{x^2} + bx + c = 0$, the sum of the roots of the quadratic equation is given by $\dfrac{{ - b}}{a}$ and the product of the roots of the quadratic equation is given by $\dfrac{c}{a}$.
Complete step by step Solution:
Expand the equation $\dfrac{\alpha }{{x - \alpha }} + \dfrac{\beta }{{x - \beta }} = 1$
$\dfrac{{\alpha (x - \beta ) + \beta (x - \alpha )}}{{(x - \alpha )(x - \beta )}} = 1$
$\alpha (x - \beta ) + \beta (x - \alpha ) = (x - \alpha )(x - \beta )$
Expanding further,
$\alpha x - \alpha \beta + \beta x - \alpha \beta = {x^2} - \alpha x - \beta x + \alpha \beta $
${x^2} - 2\alpha x - 2\beta x + 3\alpha \beta = 0$
${x^2} - 2(\alpha + \beta )x + 3\alpha \beta = 0$
The above equation is of the form $a{x^2} + bx + c = 0$. We know that the sum of roots of a quadratic equation of the form $a{x^2} + bx + c = 0$ is $\dfrac{{ - b}}{a}$. It is given to us that the roots are equal in magnitude and opposite in sign. Therefore, the sum of roots must equal zero.
Therefore,
\[\;2(\alpha + \beta ) = 0\]
\[\;\alpha + \beta = 0\]
Therefore, the correct option is (A).
Note:Given an equation in the standard form $a{x^2} + bx + c = 0$, the sum of the roots of the quadratic equation is given by $\dfrac{{ - b}}{a}$ and the product of the roots of the quadratic equation is given by $\dfrac{c}{a}$.
Recently Updated Pages
If there are 25 railway stations on a railway line class 11 maths JEE_Main

Minimum area of the circle which touches the parabolas class 11 maths JEE_Main

Which of the following is the empty set A x x is a class 11 maths JEE_Main

The number of ways of selecting two squares on chessboard class 11 maths JEE_Main

Find the points common to the hyperbola 25x2 9y2 2-class-11-maths-JEE_Main

A box contains 6 balls which may be all of different class 11 maths JEE_Main

Trending doubts
JEE Main 2025 Session 2: Application Form (Out), Exam Dates (Released), Eligibility, & More

JEE Main 2025: Derivation of Equation of Trajectory in Physics

Electric Field Due to Uniformly Charged Ring for JEE Main 2025 - Formula and Derivation

Displacement-Time Graph and Velocity-Time Graph for JEE

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

JEE Main Eligibility Criteria 2025

Other Pages
JEE Advanced Marks vs Ranks 2025: Understanding Category-wise Qualifying Marks and Previous Year Cut-offs

JEE Advanced 2025: Dates, Registration, Syllabus, Eligibility Criteria and More

JEE Advanced Weightage 2025 Chapter-Wise for Physics, Maths and Chemistry

NCERT Solutions for Class 11 Maths Chapter 4 Complex Numbers and Quadratic Equations

NCERT Solutions for Class 11 Maths In Hindi Chapter 1 Sets

NCERT Solutions for Class 11 Maths Chapter 6 Permutations and Combinations
