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Mathematics 59 Online
OpenStudy (anonymous):

Please check this... Have I done it right?? Can anybody suggest any other method to solve this: If \(x = 3 + 2\sqrt{2}\) Then, find: \[\sqrt{x} - \frac{1}{\sqrt{x}} = \]

OpenStudy (anonymous):

It will take time so you guys can do other work.. I am just writing..

OpenStudy (zepp):

I'm writing too.

OpenStudy (anonymous):

Given: \(x = 3 + 2 \sqrt{2}\) I have just did like this: Squaring it: \[(\sqrt{x} - \frac{1}{\sqrt{x} })^2 = x + \frac{1}{x} - 2\] \[= \frac{x^2 + 1 - 2x}{x} = \frac{(x-1)^2}{x}\] Now : \[(x - 1) = 2 + 2\sqrt{2}\] \[(x-1)^2 = 12 + 8\sqrt{2}\] Put in the expression above: \[= \frac{12 + 8\sqrt{2}}{3 + 2\sqrt{2}} = \frac{12 + 8\sqrt{2}}{3 + 2\sqrt{2}} \times \frac{(3 - 2\sqrt{2})}{(3 - 2\sqrt{2})}\] \[= 36 - 24\cancel{\sqrt{2}} + \cancel{24\sqrt{2}} - 32 = 4\] Taking square root both the sides: \[\sqrt{x} - \frac{1}{\sqrt{x}} = \sqrt{4} = 2\]

OpenStudy (vishweshshrimali5):

but sq. root can have both + and - ve values right ?

OpenStudy (anonymous):

Yeah right.. But I am not asking about the answer but the solution.. and of course you are right in that..

OpenStudy (anonymous):

Is there any other method simpler than this??

OpenStudy (apoorvk):

Nah, this is he simplest method in the sense, that it's the least calculative and the most tactical. Nice solution there.

OpenStudy (vishweshshrimali5):

a better one ii think

OpenStudy (vishweshshrimali5):

i just calculated sqrt x and solved is it the smartest one

OpenStudy (vishweshshrimali5):

:)

OpenStudy (apoorvk):

Oh no, I didn't notice that \(3+2\sqrt2\) thingy, it's popular. Hmm, it IS the smartest procedure Vishwesh, good one!

OpenStudy (vishweshshrimali5):

Thanks @apoorvk

OpenStudy (anonymous):

Yes this is the best.. @vishweshshrimali5

OpenStudy (vishweshshrimali5):

@waterineyes thanks

OpenStudy (anonymous):

Thanks for what??? Even, I should say thanks to you.. Thanks @vishweshshrimali5

OpenStudy (vishweshshrimali5):

:)

OpenStudy (apoorvk):

Yeah don't thank us Vishwesh, we should thank you for that brilliant work ;)

OpenStudy (vishweshshrimali5):

:)

mathslover (mathslover):

thanks a lot every one for compliments :D gr8 work @vishweshshrimali5

OpenStudy (vishweshshrimali5):

:)

OpenStudy (anonymous):

This is called Quality Solution to this problem..

mathslover (mathslover):

thanks again :)

OpenStudy (anonymous):

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