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# Mehul's doubt.

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

$\dfrac{a+2}{a-2} - \dfrac{a+2\sqrt{2}}{a-2\sqrt{2}}$

Note by Nihar Mahajan
2 years ago

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

Err, sir, There is a typo in the line of the answer. It should be "+" · 2 years ago

Thanks. · 2 years ago

$$\huge\ddot\smile$$ · 1 year, 12 months ago

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

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. · 2 years ago

No need to @Mention!!! $$\huge \ddot \smile$$ · 2 years ago

Yes , I am correct then. :P · 2 years ago

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

Simply put the value of $$a$$ and calculate. :P · 2 years ago

12+10\sqrt{2} · 2 years ago

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

Can u tell me the options @Mehul Arora · 2 years ago

Sure. a. $$12+ \sqrt {2}$$

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

c. 2

d. -2 · 2 years ago

According to the question, we have to find,

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

You have made a mistake.Find it. · 2 years ago

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

@Mehul Arora , I am afraid that there is no such option in the paper which is the correct value of the expression above. · 2 years ago

Thanks! I just got juggled up with the signs.

$$cheers!$$ · 2 years ago

There should be one more option as "None of the given". :P · 2 years ago

Yeah... I calculated it manually. :P · 2 years ago

ha ha componendo dividendo solved it for me isnt that way easier??? · 1 year, 5 months ago

How did you solve this using componendo dividendo ? Also , post your solution. · 1 year, 5 months ago

sorry i cant dont know a thing about latex. · 1 year, 5 months ago
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