HongKong 2022-23 IMO Selection Tests

Let $u=\sqrt{2x+3}$. Then the equation becomes $2x^2+3u^2=7xu$, or $(2x-u)(x-3u)=0$. Hence either $u=2x$ or $x=3u$.

* If $u=2x$, we get $2x=\sqrt{2x+3}$. Squaring both sides gives $4x^2=2x+3$, which leads to $x=\frac{1\pm \sqrt{13}}{4}$.

* If $x=3u$, we get $\frac{x}{3}=\sqrt{2x+3}$. Squaring both sides gives $\frac{1}{9}{x}^2=2x+3$, which leads to $x=9\pm 6\sqrt{3}$.

Among these, the largest one is $x=9+6\sqrt{3}$. One can check that in this case $$3u=\sqrt{18x+27}=\sqrt{189+108\sqrt{3}}=\sqrt{(9+\sqrt{108}{)}^2}=x$$ and so such $x$ satisfies the original equation, which is therefore the largest real root of the equation.