Question

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 1

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 $\dfrac{c}{a}$.