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sqrt{(2 sqrt 63) +2/8+3 sqrt 7}

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by bhumika.k.shah » Thu Jan 21, 2010 1:48 am
how to go about with this?


OA D
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by ajith » Thu Jan 21, 2010 2:29 am
bhumika.k.shah wrote:how to go about with this?


OA D



√(2√63+ 2/(8+3√7)) = √(2√63+ 2/(8+√63))= √(126+16√63+ 2)/(8+√63)) = √16((8+√63)/ (8+√63)) =√16 =4
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by bhumika.k.shah » Thu Jan 21, 2010 2:33 am
how did u get the third step???
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by ajith » Thu Jan 21, 2010 2:42 am
bhumika.k.shah wrote:how did u get the third step???
1/(a+b) + c = (ab+ac+1)/(a+b)

so,

(2√63+ 2/(8+√63))= (2√63*8 + 2√63* √63 + 2) /(8+√63) = (16√63 + 2*63 + 2)/(8+√63) = (128+16√63)/(8+√63)
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by bhumika.k.shah » Thu Jan 21, 2010 2:45 am
any other approach????
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by linkinpark » Thu Jan 21, 2010 6:44 am
sqrt(2*sqrt63 + 2/(8+3*sqrt(3))
=> sqrt(2*sqrt63 + (2*(8 - 3*sqrt(3))/1 (multiply/divide by 8 - 3*sqrt(3) )
=> sqrt(2*sqrt63 + 16 - 2*sqrt63) = sqrt(16) = 4
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by bhumika.k.shah » Thu Jan 21, 2010 7:07 am
linkin park ,

i think u got the question wrong...its 8+3*sqrt7
linkinpark wrote:sqrt(2*sqrt63 + 2/(8+3*sqrt(3))
=> sqrt(2*sqrt63 + (2*(8 - 3*sqrt(3))/1 (multiply/divide by 8 - 3*sqrt(3) )
=> sqrt(2*sqrt63 + 16 - 2*sqrt63) = sqrt(16) = 4
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by linkinpark » Thu Jan 21, 2010 7:09 am
sqrt(2*sqrt63 + 2/(8+3*sqrt(7))
=> sqrt(2*sqrt63 + (2*(8 - 3*sqrt(7))/1 (multiply/divide by 8 - 3*sqrt(7) )
=> sqrt(2*sqrt63 + 16 - 2*sqrt63) = sqrt(16) = 4

*corrected the numbers, Bhumika thanks for pointing numbers were wrong but solution stays
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by bhumika.k.shah » Thu Jan 21, 2010 7:11 am
sowree but could u please elaborate these two points??

(2*sqrt63 + 16 - 2*sqrt63) =

sqrt(16) = 4

Thanks :)
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by linkinpark » Thu Jan 21, 2010 7:24 am
square_root[2*square_root(63) + 16 - 2*square_root(63) ]
will give me square_root(16) because other two terms are of opposite sign and of equal value so both cancel out

HTH
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