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Q.
The number of real roots of $\left(x+\frac{1}{x}\right)^{3} +\left(x+\frac{1}{x}\right)=0$ is
Complex Numbers and Quadratic Equations
Solution:
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.