Question

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

• A. $0$
• B. $2$
• C. $4$
• D. $6$

Answer 1

Answer: (A)

We have, ${\left(x+\frac{1}{x}\right)}^3+\left(x+\frac{1}{x}\right)=0$
$\Rightarrow \left(x+\frac{1}{x}\right)\left[{\left(x+\frac{1}{x}\right)}^2+1\right]=0$
$\Rightarrow$ either $x+\frac{1}{x}=0$
$\Rightarrow x^2=-1$
$\Rightarrow x=\pm i$
or ${\left(x+\frac{1}{x}\right)}^2+1=0$
$\Rightarrow x^2+\frac{1}{x^2}+3=0$
$\Rightarrow x^4+3x^2+1=0$
$\Rightarrow x^2=\frac{-3\pm \sqrt{9-4}}{2}$
$=\frac{-3\pm \sqrt{5}}{2}<0$
$\therefore$ There is no real root.