Question

The number of real roots of the equation ${\left( {x + \left( {\dfrac{1}{x}} \right)} \right)^3} + x + \left( {\dfrac{1}{x}} \right) = 0$ is
A. $0$
B. $2$
C. $4$
D. $6$

Answer 1

Hint: In this question, we are given an equation ${\left( {x + \left( {\dfrac{1}{x}} \right)} \right)^3} + x + \left( {\dfrac{1}{x}} \right) = 0$ and we have to find the roots or can say we have to check how many roots are real. First step is to take $x + \left( {\dfrac{1}{x}} \right)$ common from the given equation. You’ll get two cases in which one cannot be possible. So, find the roots using another case.

Complete step by step solution: 
Given that,
${\left( {x + \left( {\dfrac{1}{x}} \right)} \right)^3} + x + \left( {\dfrac{1}{x}} \right) = 0$
Taking $x + \left( {\dfrac{1}{x}} \right)$ common in the above equation,
$\left( {x + \left( {\dfrac{1}{x}} \right)} \right)\left( {{{\left( {x + \left( {\dfrac{1}{x}} \right)} \right)}^2} + 1} \right) = 0$
Here, this term $\left( {{{\left( {x + \left( {\dfrac{1}{x}} \right)} \right)}^2} + 1} \right)$ is greater than to $0$and cannot be equal to zero.
It implies that, $x + \left( {\dfrac{1}{x}} \right) = 0$
${x^2} + 1 = 0$
${x^2} = - 1$
$x = \pm i$
The required roots are imaginary.
So, Option (A) is the correct answer..

Note: The key concept involved in solving this problem is the good knowledge of Real and imaginary roots. Students must remember that the primary distinction between real and complex roots is that real roots are expressed as real numbers, while complex roots are expressed as imaginary numbers. Imaginary roots are expressed in imaginary numbers, with $i = \sqrt { - 1} $ being the simplest imaginary number. Most imaginary numbers can be expressed as $a + ib$ where $a$ and $b$ are real numbers but the whole number is imaginary due to the presence of $i$. When we solve the equation by taking the square root of a negative number, we get imaginary roots.