Answer
There are no real solutions.
Work Step by Step
$f\left( x \right)={{\left( \frac{{{x}^{2}}+1}{x} \right)}^{4}}+4{{\left( \frac{{{x}^{2}}+1}{x} \right)}^{2}}+12$.
The x-intercept occurs when$f\left( x \right)=0$,
${{\left( \frac{{{x}^{2}}+1}{x} \right)}^{4}}+4{{\left( \frac{{{x}^{2}}+1}{x} \right)}^{2}}+12=0$ …… (1)
Let $u={{\left( \frac{{{x}^{2}}+1}{x} \right)}^{2}}$ and${{u}^{2}}={{\left( \frac{{{x}^{2}}+1}{x} \right)}^{4}}$,
Substitute the values of $u$ and ${{u}^{2}}$ in equation (1),
${{u}^{2}}+4u+12=0$
$\begin{align}
& x=\frac{-\left( 4 \right)\pm \sqrt{{{\left( 4 \right)}^{2}}-4\left( 1 \right)\left( 12 \right)}}{2\left( 1 \right)} \\
& =\frac{-7\pm \sqrt{16-48}}{2} \\
& =\frac{-7\pm \sqrt{-32}}{2}
\end{align}$
There are no real solutions.
