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.
 A: $$x(5+2\sqrt{6})=289\\
\Rightarrow x=\frac{289}{(5+2\sqrt{6})}\\
\Rightarrow x=\frac{289(5-2\sqrt{6})}{(5+2\sqrt{6})(5-2\sqrt{6})}\\
\Rightarrow x=\frac{289(5-2\sqrt{6})}{25-24}\\
\Rightarrow x=-289(2\sqrt{6}-5)\\$$
A: We have
$$x(5+2\sqrt{6})=289$$
$$\Rightarrow x=\frac{289}{5+2\sqrt{6}}$$
$$\Rightarrow x=\frac{289}{5+2\sqrt{6}}\frac{5-2\sqrt{6}}{5-2\sqrt{6}}$$
$$\Rightarrow x=\frac{289(5-2\sqrt{6})}{25-24}$$
$$\Rightarrow x=289(5-2\sqrt{6})$$
A: I think your question is about $\sqrt{x}(\sqrt{3}+\sqrt{2})=17$. Then
$$
\begin{array}{rcl}
\sqrt{x}(\sqrt{3}+\sqrt{2}) & = & 17 \\
(\sqrt{x}(\sqrt{3}+\sqrt{2}))^2 & = & 17^2 \\
x(\sqrt{3}^2+2\sqrt{3}\sqrt{2}+\sqrt{2}^2) & = & 289 \\
x(3+2\sqrt{3}\sqrt{2}+2) & = & 289 \\
x(5+2\sqrt{3\times2}) & = & 289\\
x(5+2\sqrt{6}) & = & 289 \\
x & = & \frac{289}{5+2\sqrt{6}} \\
x & = & \frac{289}{5+2\sqrt{6}} \times \frac{5-2\sqrt{6}}{5-2\sqrt{6}}  \\
x & = & \frac{289 \times (5-2\sqrt{6})}{5^2+4\times 6} \\
x & = & \frac{289 \times (5-2\sqrt{6})}{5^2-4\times 6} \\
x & = & \frac{289 \times (5-2\sqrt{6})}{25 - 24} \\
x & = & 289 \times (5-2\sqrt{6})
\end{array}
$$
A: $$(\sqrt{3x}+\sqrt{2x})^2=17^2=289$$
$$(\sqrt{3x}+\sqrt{2x})^2 = 3x + 2x + 2\sqrt{6}x= (5+2\sqrt{6})x=289,$$
hence 
$$x=\frac{289}{(5+2\sqrt{6})} = \frac{289 (5-2\sqrt{6})}{5^2-(2\sqrt{6})^2}
=289 (5-2\sqrt{6})$$
A: I think $$x = \left( \frac{17}{\sqrt{3} + \sqrt{2}} \right)^2$$ is a perfectly fine solution, absent any contextual reason to rationalize the denominator.
