<s>[INST] Solve $\sqrt{3x}+\sqrt{2x}=17$ This is what I did:
$$\sqrt{3x}+\sqrt{2x}=17$$
$$\implies\sqrt{3x}+\sqrt{2x}=17$$
$$\implies\sqrt{3}\sqrt{x}+\sqrt{2}\sqrt{x}=17$$
$$\implies\sqrt{x}(\sqrt{3}+\sqrt{2})=17$$
$$\implies x(5+2\sqrt{6})=289$$
I don't know how to continue. And when I went to wolfram alpha, I got:
$$x=-289(2\sqrt{6}-5)$$
Could you show me the steps to get the final result?
Thank you.
 [/INST] We have
$$x(5+2\sqrt{6})=289$$
$$\Rightarrow x=\frac{289}{5+2\sqrt{6}}$$
$$\Rightarrow x=\frac{289}{5+2\sqrt{6}}\frac{5-2\sqrt{6}}{5-2\sqrt{6}}$$
$$\Rightarrow x=\frac{289(5-2\sqrt{6})}{25-24}$$
$$\Rightarrow x=289(5-2\sqrt{6})$$
</s>