The substitution $3x-1=t^2$ it work fine. After this substitution we get the equation ${{t}^{4}}+2{{t}^{2}}-27t+10=0$or $({{t}^{2}}-3t+1)({{t}^{2}}+3t+10)=0$and the rest it is easy.

p.lazarov06 wrote: Solve the equation: \[3\sqrt{3x-1}=x^2+1\] Since both sides are $\ge 0$, equivalent to square and we get $x^4+2x^2-27x+10=0$ Which is $(x^2-3x+1)(x^2+3x+10)=0$ And so $\boxed{x\in\left\{\frac{3-\sqrt 5}2,\frac{3+\sqrt 5}2\right\}}$

p.lazarov06 wrote: Solve the equation: \[3\sqrt{3x-1}=x^2+1\] Other approach : We need $x\ge \frac 13$ and we have $x^2-3x+1=3(\sqrt{3x-1}-x)=-3\frac{x^2-3x+1}{x+\sqrt{3x-1}}$ and so $x^2-3x+1=0$ (since $x+\sqrt{3x-1}>0$)