Weird coefficient

$\begin{aligned} \sqrt{x^3+3x^2+9x-13} & = \sqrt{3-2x}+\frac{\sqrt{543}-\sqrt{27}}{9} \\ \sqrt{x^3+3x^2+9x-13} - \sqrt{3-2x} & =\sqrt{\frac {181}{27}}-\sqrt{\frac13} \end{aligned}$

Equating the radicals and consider the smaller ones first.

$\begin{aligned} \sqrt{3-2x} & =\sqrt{\frac13} \\ \implies 3-2x & = \frac13 \\ 2x & = \frac 83 \\ \implies x & = \frac 43 \end{aligned}$

Substituting $x = \frac 43$ is the first radical,

$\begin{aligned} \sqrt {x^3+3x^2+9x-13} & = \sqrt {\frac {64}{27} +3 \cdot \frac {16}9 + 9 \cdot \frac 43-13} \\ & = \sqrt {\frac {64+144+324-351}{27}} \\ & = \sqrt {\frac {181}{27}} = RHS \end{aligned}$

Note that the equation is only valid for $1 \le x \le \frac 32$ , because $x^3+3x^2+9x-13< 0$ for $x < 1$ and $3-2x < 0$ for $x > \frac 32$ . Also note that within $x \in \left[1, \frac 32\right]$ , $x^3+3x^2+9x-13$ is continuously increasing and $3-2x$ , continuously decreasing, therefore, $x = \frac 43$ is the only real root and that the sum of real roots is also $\frac 43 \approx \boxed{1.33}$ .

Here's my approach. The constrain of $x$ is $x\in [1;\frac{3}{2}]$ $\sqrt{x^3+3x^2+9x-13}=\sqrt{3-2x}+\frac{\sqrt{543}-\sqrt{27}}{9}$ $\Leftrightarrow \sqrt{x^3+3x^2+9x-13}-\frac{\sqrt{543}}{9}+\frac{\sqrt{27}}{9}-\sqrt{3-2x}=0$ $\Leftrightarrow\frac{x^3-3x^2+9x-\frac{532}{27}}{\sqrt{x^3+3x^2+9x-13}+\frac{\sqrt{543}}{9}}+\frac{2x-\frac{8}{3}}{\sqrt{3-2x}+\frac{\sqrt{27}}{9}}=0$ $\Leftrightarrow \big(x-\frac{4}{3}\big)\bigg(\frac{x^2+\frac{13x}{3}+\frac{133}{9}}{\sqrt{x^3+3x^2+9x-13}+\frac{\sqrt{543}}{9}}+\frac{2}{\sqrt{3-2x}+\frac{\sqrt{27}}{9}}\bigg)=0$ We see that $\frac{x^2+\frac{13x}{3}+\frac{133}{9}}{\sqrt{x^3+3x^2+9x-13}+\frac{\sqrt{543}}{9}}+\frac{2}{\sqrt{3-2x}+\frac{\sqrt{27}}{9}}>0 ; \forall x\in [1;\frac{3}{2}]$ , so the only root is $x=\frac{4}{3}\approx 1.33$