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.
$$x(5+2\sqrt{6})=289\\
\Rightarrow x=\frac{289}{(5+2\sqrt{6})}\\
\Rightarrow x=\frac{289(5-2\sqrt{6})}{(5+2\sqrt{6})(5-2\sqrt{6})}\\
\Rightarrow x=\frac{289(5-2\sqrt{6})}{25-24}\\
\Rightarrow x=-289(2\sqrt{6}-5)\\$$
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.