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This discussion board is a place to discuss our Daily Challenges and the math and science related to those challenges. Explanations are more than just a solution — they should explain the steps and thinking strategies that you used to obtain the solution. Comments should further the discussion of math and science.

When posting on Brilliant:

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## Comments

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TopNewest$\begin{aligned} \dfrac{a+2}{a-2} - \dfrac{a+2\sqrt{2}}{a-2\sqrt{2}} & = \dfrac{8-4\sqrt{2}+2}{8-4\sqrt{2}-2} - \dfrac{8-4\sqrt{2}+2\sqrt{2}}{8-4\sqrt{2}-2\sqrt{2}} \\ & = \dfrac{10-4\sqrt{2}}{6-4\sqrt{2}} - \dfrac{8-2\sqrt{2}}{8-6\sqrt{2}} \\ & = \dfrac{5-2\sqrt{2}}{3-2\sqrt{2}} - \dfrac{4-\sqrt{2}}{4-3\sqrt{2}} \\ & = \dfrac{(5-2\sqrt{2})(3+2\sqrt{2})}{9-8} - \dfrac{(4-\sqrt{2})(4+3\sqrt{2})}{16-18} \\ & = 15+4\sqrt{2} - 8 + \dfrac{16 + 8\sqrt{2} - 6}{2} \\ & = 7 +4\sqrt{2} + 5 + 4\sqrt{2} \\ & = 12+8\sqrt{2} \end{aligned}$

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Err, sir, There is a typo in the line of the answer. It should be "+"

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Thanks.

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$\huge\ddot\smile$

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12+10\sqrt{2}

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@Soumya Khurana , I am afraid that there is no such option in the paper I got this from.

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Can u tell me the options @Mehul Arora

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$12+ \sqrt {2}$

Sure. a.b. $12- \sqrt {2}$

c. 2

d. -2

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@Mehul Arora , I am afraid that there is no such option in the paper which is the correct value of the expression above.

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$cheers!$

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@Mehul Arora I checked for sure using a claculator...None of the options match. :(

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According to the question, we have to find,

$\huge \dfrac{8-4\sqrt{2}+2}{8-4\sqrt{2}-2}-\dfrac{8-4\sqrt{2}+2\sqrt{2}}{8-4\sqrt{2}-2\sqrt{2}} \\ \huge =\dfrac{10-4\sqrt{2}}{6-4\sqrt{2}} \times \dfrac{6+4\sqrt{2}}{6+\sqrt{2}} - \dfrac{8-2\sqrt{2}}{8-6\sqrt{2}} \times \dfrac{8+6\sqrt{2}}{8+6\sqrt{2}} \\ \huge = 7+4\sqrt{2}-(-5-4\sqrt{2}) \\ \huge = \boxed{12+8\sqrt{2}}$

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Simply put the value of $a$ and calculate. :P

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The answer I am getting is $12 + 8\sqrt{2}$ , Is it correct?

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The correct ans is $12+8\sqrt{2}$.

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Yes , I am correct then. :P

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Yeah. Thank you sir. :)

I needed the answer. I found out the procedure.

Thanks everyone @Nihar Mahajan @Sravanth Chebrolu @Soumya Khurana and of course @Sandeep Bhardwaj sir.

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No need to @Mention!!! $\huge \ddot \smile$

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ha ha componendo dividendo solved it for me isnt that way easier???

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How did you solve this using componendo dividendo ? Also , post your solution.

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sorry i cant dont know a thing about latex.

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