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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.

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    $\begingroup$ That is the final result. Perhaps you should rationalise it. $\endgroup$ – ajotatxe Oct 16 '14 at 22:59
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    $\begingroup$ what is the reciprocal of $5 + 2 \sqrt 6?$ The usual phrase for this would be "rationalize the denominator" of $$ \frac{1}{5 + 2 \sqrt 6} $$ $\endgroup$ – Will Jagy Oct 16 '14 at 23:00
  • $\begingroup$ @ajotatxe I think this was targeted at the OP... $\endgroup$ – user155385 Oct 16 '14 at 23:03
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    $\begingroup$ @ajotatxe, yes, I wanted Mathink to do it and figure out how to use that to finish the problem. I ought to have written Mathink as the first word $\endgroup$ – Will Jagy Oct 16 '14 at 23:03
  • $\begingroup$ Yes... too much math seems to make me a bit stupid... xD $\endgroup$ – ajotatxe Oct 16 '14 at 23:06
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$$x(5+2\sqrt{6})=289\\ \Rightarrow x=\frac{289}{(5+2\sqrt{6})}\\ \Rightarrow x=\frac{289(5-2\sqrt{6})}{(5+2\sqrt{6})(5-2\sqrt{6})}\\ \Rightarrow x=\frac{289(5-2\sqrt{6})}{25-24}\\ \Rightarrow x=-289(2\sqrt{6}-5)\\$$

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We have

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

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

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$$(\sqrt{3x}+\sqrt{2x})^2=17^2=289$$ $$(\sqrt{3x}+\sqrt{2x})^2 = 3x + 2x + 2\sqrt{6}x= (5+2\sqrt{6})x=289,$$ 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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  • $\begingroup$ Thank you very much,now i got it. And many thanks to everyone that helped me. $\endgroup$ – Mathink Oct 16 '14 at 23:03
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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.

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  • $\begingroup$ You guys, you're great. Thank you so much. $\endgroup$ – Mathink Oct 16 '14 at 23:09

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