Thank you for reporting, we will resolve it shortly
Q.
The number of real roots of
$(x - 1) (x - 2) (x - 3) (x - 4) = 3$ is
Complex Numbers and Quadratic Equations
Solution:
$
\begin{array}{l}
\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]
\end{array}
$
Since, $\left(x+\frac{1}{x}\right)^{2}+1>0$
Therefore, $\left(x+\frac{1}{x}\right)=0$
$
\Rightarrow x ^{2}+1=0
$
Therefore, number of real roots are zero.