Question

If \[a,b,c,\alpha ,\beta ,\gamma ,\delta \] are real then number of imaginary roots of $\dfrac{{{a^2}}}{{x - \alpha }} + \dfrac{{{b^2}}}{{x - \beta }} + \dfrac{{{c^2}}}{{x - \gamma }} - x + \delta = 0$ .
$\left( A \right){\text{ 1}}$
$\left( B \right){\text{ 0}}$
$\left( C \right){\text{ 3}}$
$\left( D \right){\text{ 2}}$

Answer 1

Hint:
As we know that the when the \[a,b,c,\alpha ,\beta ,\gamma ,\delta \] are all real then we will take $a = b = c = 1$ and \[\alpha = \beta = \gamma = \delta = 0\]. So by substituting the values, we will get the value for the $x$ and in this way we can answer this question.

Complete step by step solution:
It is given that the \[a,b,c,\alpha ,\beta ,\gamma ,\delta \] are real so from this we know that when \[a,b,c,\alpha ,\beta ,\gamma ,\delta \] is real then $a = b = c = 1$ and \[\alpha = \beta = \gamma = \delta = 0\] .
So we will substitute the values, in the equation $\dfrac{{{a^2}}}{{x - \alpha }} + \dfrac{{{b^2}}}{{x - \beta }} + \dfrac{{{c^2}}}{{x - \gamma }} - x + \delta = 0$ , we get
$ \Rightarrow \dfrac{{{1^2}}}{{x - 0}} + \dfrac{{{1^2}}}{{x - 0}} + \dfrac{{{1^2}}}{{x - 0}} - x + 0 = 0$
Now on solving the above equation, we get
$ \Rightarrow \dfrac{1}{x} + \dfrac{1}{x} + \dfrac{1}{x} - x = 0$
Now we will take the LCM of the denominator and will solve the above equation, we get
$ \Rightarrow \dfrac{{1 + 1 + 1}}{x} - x = 0$
So on solving the above equation, we get
$ \Rightarrow \dfrac{3}{x} - x = 0$
Now for solving this, we will take the LCM first and then we will solve the subtraction, we get
$ \Rightarrow \dfrac{{3 - {x^2}}}{x} = 0$
Now taking the denominator to the right side of the equation, so we will get the equation as
$ \Rightarrow {x^2} - 3 = 0$
So, we will solve for the value of $x$ and for this we will take the constant term to the right side of the equation, therefore the equation will be
$ \Rightarrow {x^2} = 3$
On removing the square, we get
$ \Rightarrow x = \pm \sqrt 3 $
Since, there is no complex roots and
Therefore, the number of imaginary roots for the equation will be zero.

Hence, the option $\left( 2 \right)$ is correct.

Note:
On a note, we should know that if the roots of the equation are imaginary then it will always occur in conjugate. That is the imaginary and the complex roots will occur and so if it will not be in the equation then we will call it zero and so as in question we had called it.