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Question: If \[a,b,c,\alpha ,\beta ,\gamma ,\delta \] are real then number of imaginary roots of \(\dfrac{{{a^...

If a,b,c,α,β,γ,δa,b,c,\alpha ,\beta ,\gamma ,\delta are real then number of imaginary roots of a2xα+b2xβ+c2xγx+δ=0\dfrac{{{a^2}}}{{x - \alpha }} + \dfrac{{{b^2}}}{{x - \beta }} + \dfrac{{{c^2}}}{{x - \gamma }} - x + \delta = 0 .
(A) 1\left( A \right){\text{ 1}}
(B) 0\left( B \right){\text{ 0}}
(C) 3\left( C \right){\text{ 3}}
(D) 2\left( D \right){\text{ 2}}

Explanation

Solution

As we know that the when the a,b,c,α,β,γ,δa,b,c,\alpha ,\beta ,\gamma ,\delta are all real then we will take a=b=c=1a = b = c = 1 and α=β=γ=δ=0\alpha = \beta = \gamma = \delta = 0. So by substituting the values, we will get the value for the xx and in this way we can answer this question.

Complete step by step solution:
It is given that the a,b,c,α,β,γ,δa,b,c,\alpha ,\beta ,\gamma ,\delta are real so from this we know that when a,b,c,α,β,γ,δa,b,c,\alpha ,\beta ,\gamma ,\delta is real then a=b=c=1a = b = c = 1 and α=β=γ=δ=0\alpha = \beta = \gamma = \delta = 0 .
So we will substitute the values, in the equation a2xα+b2xβ+c2xγx+δ=0\dfrac{{{a^2}}}{{x - \alpha }} + \dfrac{{{b^2}}}{{x - \beta }} + \dfrac{{{c^2}}}{{x - \gamma }} - x + \delta = 0 , we get
12x0+12x0+12x0x+0=0\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
1x+1x+1xx=0\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
1+1+1xx=0\Rightarrow \dfrac{{1 + 1 + 1}}{x} - x = 0
So on solving the above equation, we get
3xx=0\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
3x2x=0\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
x23=0\Rightarrow {x^2} - 3 = 0
So, we will solve for the value of xx and for this we will take the constant term to the right side of the equation, therefore the equation will be
x2=3\Rightarrow {x^2} = 3
On removing the square, we get
x=±3\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 (2)\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.