New user? Sign up

Existing user? Log in

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 4 years, 1 month ago

$</code> ... <code>$</code>...<code>."> Easy Math Editor

*italics*

_italics_

**bold**

__bold__

- bulleted- list

1. numbered2. list

paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)

> This is a quote

This is a quote

# I indented these lines # 4 spaces, and now they show # up as a code block. print "hello world"

2 \times 3

2^{34}

a_{i-1}

\frac{2}{3}

\sqrt{2}

\sum_{i=1}^3

\sin \theta

\boxed{123}

Sort by:

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

Log in to reply

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

Thanks.

@Chew-Seong Cheong – $\huge\ddot\smile$

12+10\sqrt{2}

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

Can u tell me the options @Mehul Arora

@Soumya Khurana – Sure. a. $12+ \sqrt {2}$

b. $12- \sqrt {2}$

c. 2

d. -2

@Mehul Arora – @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 – There should be one more option as "None of the given". :P

@Sandeep Bhardwaj – Yeah... I calculated it manually. :P

@Nihar Mahajan – Thanks! I just got juggled up with the signs.

$cheers!$

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

@Mehul Arora – Please see the explanation,

According to the question, we have to find,

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

@Sravanth Chebrolu – You have made a mistake.Find it.

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

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

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

Yes , I am correct then. :P

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.

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

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

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

sorry i cant dont know a thing about latex.

@Kaustubh Miglani – You can post a solution without latex too. I will understand it. Or at least post your approach.I am curious about this method.

Problem Loading...

Note Loading...

Set Loading...

$</code> ... <code>$</code>...<code>."> Easy Math Editor

`*italics*`

or`_italics_`

italics`**bold**`

or`__bold__`

boldNote: you must add a full line of space before and after lists for them to show up correctlyparagraph 1

paragraph 2

`[example link](https://brilliant.org)`

`> This is a quote`

Remember to wrap math in $</span> ... <span>$ or $</span> ... <span>$ to ensure proper formatting.`2 \times 3`

`2^{34}`

`a_{i-1}`

`\frac{2}{3}`

`\sqrt{2}`

`\sum_{i=1}^3`

`\sin \theta`

`\boxed{123}`

## Comments

Sort by:

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

Log in to reply

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

Log in to reply

Thanks.

Log in to reply

$\huge\ddot\smile$

Log in to reply

12+10\sqrt{2}

Log in to reply

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

Log in to reply

Can u tell me the options @Mehul Arora

Log in to reply

$12+ \sqrt {2}$

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

c. 2

d. -2

Log in to reply

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

Log in to reply

Log in to reply

Log in to reply

$cheers!$

Log in to reply

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

Log in to reply

According to the question, we have to find,

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

Log in to reply

Log in to reply

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

Log in to reply

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

Log in to reply

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

Log in to reply

Yes , I am correct then. :P

Log in to reply

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.

Log in to reply

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

Log in to reply

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

Log in to reply

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

Log in to reply

sorry i cant dont know a thing about latex.

Log in to reply

Log in to reply