The number of real solutions of the equation $3\left({\mathrm{x}}^2+\frac{1}{{\mathrm{x}}^2}\right)-2\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)+5=0$ is

$3\left[{\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)}^2-2\right]-2\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)+5=0$
Put $\mathrm{x}+\frac{1}{\mathrm{x}}=\mathrm{y}$
$\begin{array}{l}3\left({\mathrm{y}}^2-2\right)-2\mathrm{y}+5=0\\ 3{\mathrm{y}}^2-2\mathrm{y}-1=0\\ 3{\mathrm{y}}^2-2\mathrm{y}+1=2\Rightarrow (3\mathrm{y}-1{)}^2=2\\ 3\mathrm{y}-1=\pm \sqrt{2}\\ 3\left(\mathrm{x}+\frac{1}{\mathrm{x}}\right)-1\pm \sqrt{2}=0\\ 3\left({\mathrm{x}}^2+1\right)+\mathrm{x}(-1\pm \sqrt{2})=0\Rightarrow 3{\mathrm{x}}^2+(-1\pm \sqrt{2})\mathrm{x}+3=0\\ 3{\mathrm{x}}^2-(1+\sqrt{2})\mathrm{x}+3=0\end{array}$

$\begin{array}{l}3{\mathrm{x}}^2+(\sqrt{2}-1)\mathrm{x}+3=0\\ {\mathrm{D}}_1=(1+\sqrt{2}{)}^{2}4(3\left)\right(3)<0,{\mathrm{D}}_2=(\sqrt{2}-1{)}^2-4\left(3\right)\left(3\right)<0\end{array}$
No, of real roots = 0