<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] I think your question is about $\sqrt{x}(\sqrt{3}+\sqrt{2})=17$. Then
$$
\begin{array}{rcl}
\sqrt{x}(\sqrt{3}+\sqrt{2}) & = & 17 \\
(\sqrt{x}(\sqrt{3}+\sqrt{2}))^2 & = & 17^2 \\
x(\sqrt{3}^2+2\sqrt{3}\sqrt{2}+\sqrt{2}^2) & = & 289 \\
x(3+2\sqrt{3}\sqrt{2}+2) & = & 289 \\
x(5+2\sqrt{3\times2}) & = & 289\\
x(5+2\sqrt{6}) & = & 289 \\
x & = & \frac{289}{5+2\sqrt{6}} \\
x & = & \frac{289}{5+2\sqrt{6}} \times \frac{5-2\sqrt{6}}{5-2\sqrt{6}}  \\
x & = & \frac{289 \times (5-2\sqrt{6})}{5^2+4\times 6} \\
x & = & \frac{289 \times (5-2\sqrt{6})}{5^2-4\times 6} \\
x & = & \frac{289 \times (5-2\sqrt{6})}{25 - 24} \\
x & = & 289 \times (5-2\sqrt{6})
\end{array}
$$
</s>