Question

The number of real roots of the equation ${\left(x+\frac{1}{x}\right)}^3+\left(x+\frac{1}{x}\right)=0$ is

• A. $0$
• B. $2$
• C. $4$
• D. $6$

Answer 1

The correct option is A

$0$

Explanation for the correct option:

Find the number of solutions of the given equation.

An equation ${\left(x+\frac{1}{x}\right)}^3+\left(x+\frac{1}{x}\right)=0$ is given.

Solve the given equation as follows:

$\left(x+\frac{1}{x}\right)\left[{\left(x+\frac{1}{x}\right)}^2+1\right]=0$

Since, ${\left(x+\frac{1}{x}\right)}^2+1\ne 0$.

So,

$$\begin{gathered} \left(x+\frac{1}{x}\right)=0 \\ \Rightarrow x^2+1=0 \\ \Rightarrow x^2=-1 \\ \Rightarrow x=\pm i \end{gathered}$$

Therefore, there is no real solution of the given equation.

Hence, option $A$ is the correct option.