If \(a=8-4\sqrt{2}\) then find the value of :
\[ \dfrac{a+2}{a-2} - \dfrac{a+2\sqrt{2}}{a-2\sqrt{2}}\]
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Nihar Mahajan
3 years, 5 months ago
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Top Newest\(\begin{equation} \begin{split} \dfrac{a+2}{a-2} - \dfrac{a+2\sqrt{2}}{a-2\sqrt{2}} & = \dfrac{8-4\sqrt{2}+2}{8-4\sqrt{2}-2} - \dfrac{8-4\sqrt{2}+2\sqrt{2}}{8-4\sqrt{2}-2\sqrt{2}} \\ & = \dfrac{10-4\sqrt{2}}{6-4\sqrt{2}} - \dfrac{8-2\sqrt{2}}{8-6\sqrt{2}} \\ & = \dfrac{5-2\sqrt{2}}{3-2\sqrt{2}} - \dfrac{4-\sqrt{2}}{4-3\sqrt{2}} \\ & = \dfrac{(5-2\sqrt{2})(3+2\sqrt{2})}{9-8} - \dfrac{(4-\sqrt{2})(4+3\sqrt{2})}{16-18} \\ & = 15+4\sqrt{2} - 8 + \dfrac{16 + 8\sqrt{2} - 6}{2} \\ & = 7 +4\sqrt{2} + 5 + 4\sqrt{2} \\ & = 12+8\sqrt{2} \end{split} \end{equation} \)
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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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The correct ans is \(12+8\sqrt{2}\).
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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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Yes , I am correct then. :P
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The answer I am getting is \(12 + 8\sqrt{2}\) , Is it correct?
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Simply put the value of \(a\) and calculate. :P
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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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Sure. a. \(12+ \sqrt {2}\)
b. \(12- \sqrt {2}\)
c. 2
d. -2
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Please see the explanation,
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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You have made a mistake.Find it.
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@Mehul Arora I checked for sure using a claculator...None of the options match. :(
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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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Thanks! I just got juggled up with the signs.
\(cheers!\)
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There should be one more option as "None of the given". :P
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Yeah... I calculated it manually. :P
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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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You can post a solution without latex too. I will understand it. Or at least post your approach.I am curious about this method.
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