<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] $$(\sqrt{3x}+\sqrt{2x})^2=17^2=289$$
$$(\sqrt{3x}+\sqrt{2x})^2 = 3x + 2x + 2\sqrt{6}x= (5+2\sqrt{6})x=289,$$
hence 
$$x=\frac{289}{(5+2\sqrt{6})} = \frac{289 (5-2\sqrt{6})}{5^2-(2\sqrt{6})^2}
=289 (5-2\sqrt{6})$$
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