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

$$\begin{gathered} 3[{(x+\frac{{\displaystyle 1}}{{\displaystyle x}})}^2-2]-2(x+\frac{{\displaystyle 1}}{{\displaystyle x}})+5=0 \\ put x+\frac{{\displaystyle 1}}{{\displaystyle x}}=y 3(y^2-2)-2y+5=0 \\ 3y^2-2y-1=0 3y^2-2y+1=2\Rightarrow {(3y-1)}^2=2 \\ 3y-1=\pm \sqrt{2} 3(x+\frac{{\displaystyle 1}}{{\displaystyle x}})-1\pm \sqrt{2}=0 \\ 3(x^2+1)+x(-1\pm \sqrt{2})=0 \Rightarrow 3x^2+(-1\pm \sqrt{2})x+3=0 \\ 3x^2-(1+\sqrt{2})x+3=0 3x^2+(\sqrt{2}-1)x+3=0 \\ {D}_1={(1+\sqrt{2})}^{2}4\left(3\right)\left(3\right)<0, D_2={(\sqrt{2}-1)}^2-4\left(3\right)\left(3\right)<0 \end{gathered}$$

No, of real roots = 0