If \(a=8-4\sqrt{2}\) then find the value of :

\[ \dfrac{a+2}{a-2} - \dfrac{a+2\sqrt{2}}{a-2\sqrt{2}}\]

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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TopNewest\(\begin{equation} \begin{split} \dfrac{a+2}{a-2} - \dfrac{a+2\sqrt{2}}{a-2\sqrt{2}} & = \dfrac{8-4\sqrt{2}+2}{8-4\sqrt{2}-2} - \dfrac{8-4\sqrt{2}+2\sqrt{2}}{8-4\sqrt{2}-2\sqrt{2}} \\ & = \dfrac{10-4\sqrt{2}}{6-4\sqrt{2}} - \dfrac{8-2\sqrt{2}}{8-6\sqrt{2}} \\ & = \dfrac{5-2\sqrt{2}}{3-2\sqrt{2}} - \dfrac{4-\sqrt{2}}{4-3\sqrt{2}} \\ & = \dfrac{(5-2\sqrt{2})(3+2\sqrt{2})}{9-8} - \dfrac{(4-\sqrt{2})(4+3\sqrt{2})}{16-18} \\ & = 15+4\sqrt{2} - 8 + \dfrac{16 + 8\sqrt{2} - 6}{2} \\ & = 7 +4\sqrt{2} + 5 + 4\sqrt{2} \\ & = 12+8\sqrt{2} \end{split} \end{equation} \) – Chew-Seong Cheong · 2 years, 2 months ago

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– Mehul Arora · 2 years, 2 months ago

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

– Chew-Seong Cheong · 2 years, 2 months ago

Thanks.Log in to reply

– Mehul Arora · 2 years, 2 months ago

\(\huge\ddot\smile\)Log in to reply

The correct ans is \(12+8\sqrt{2}\). – Sandeep Bhardwaj · 2 years, 2 months ago

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I needed the answer. I found out the procedure.

Thanks everyone @Nihar Mahajan @Sravanth Chebrolu @Soumya Khurana and of course @Sandeep Bhardwaj sir. – Mehul Arora · 2 years, 2 months ago

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– Sravanth Chebrolu · 2 years, 2 months ago

No need to @Mention!!! \(\huge \ddot \smile\)Log in to reply

– Nihar Mahajan · 2 years, 2 months ago

Yes , I am correct then. :PLog in to reply

The answer I am getting is \(12 + 8\sqrt{2}\) , Is it correct? – Nihar Mahajan · 2 years, 2 months ago

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Simply put the value of \(a\) and calculate. :P – Sandeep Bhardwaj · 2 years, 2 months ago

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12+10\sqrt{2} – Soumya Khurana · 2 years, 2 months ago

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@Soumya Khurana , I am afraid that there is no such option in the paper I got this from. – Mehul Arora · 2 years, 2 months ago

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@Mehul Arora – Soumya Khurana · 2 years, 2 months ago

Can u tell me the optionsLog in to reply

b. \(12- \sqrt {2}\)

c. 2

d. -2 – Mehul Arora · 2 years, 2 months ago

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According to the question, we have to find,

\[\huge \dfrac{8-4\sqrt{2}+2}{8-4\sqrt{2}-2}-\dfrac{8-4\sqrt{2}+2\sqrt{2}}{8-4\sqrt{2}-2\sqrt{2}} \\ \huge =\dfrac{10-4\sqrt{2}}{6-4\sqrt{2}} \times \dfrac{6+4\sqrt{2}}{6+\sqrt{2}} - \dfrac{8-2\sqrt{2}}{8-6\sqrt{2}} \times \dfrac{8+6\sqrt{2}}{8+6\sqrt{2}} \\ \huge = 7+4\sqrt{2}-(-5-4\sqrt{2}) \\ \huge = \boxed{12+8\sqrt{2}} \] – Sravanth Chebrolu · 2 years, 2 months ago

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– Nihar Mahajan · 2 years, 2 months ago

You have made a mistake.Find it.Log in to reply

@Mehul Arora I checked for sure using a claculator...None of the options match. :( – Soumya Khurana · 2 years, 2 months ago

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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. – Nihar Mahajan · 2 years, 2 months ago

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\(cheers!\) – Sravanth Chebrolu · 2 years, 2 months ago

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– Sandeep Bhardwaj · 2 years, 2 months ago

There should be one more option as "None of the given". :PLog in to reply

– Nihar Mahajan · 2 years, 2 months ago

Yeah... I calculated it manually. :PLog in to reply

ha ha componendo dividendo solved it for me isnt that way easier??? – Kaustubh Miglani · 1 year, 7 months ago

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– Nihar Mahajan · 1 year, 7 months ago

How did you solve this using componendo dividendo ? Also , post your solution.Log in to reply

– Kaustubh Miglani · 1 year, 7 months ago

sorry i cant dont know a thing about latex.Log in to reply

– Nihar Mahajan · 1 year, 7 months ago

You can post a solution without latex too. I will understand it. Or at least post your approach.I am curious about this method.Log in to reply