anonymous
  • anonymous
is my working correct? i need to find which number is bigger. i concluded that 2*sqrt38 is bigger of the two since 38^2 = 1444
Mathematics
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SOLVED
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jamiebookeater
  • jamiebookeater
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anonymous
  • anonymous
attched is my working
1 Attachment
anonymous
  • anonymous
seems correct
anonymous
  • anonymous
thanks a lot!

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anonymous
  • anonymous
(:
asnaseer
  • asnaseer
yes - correct again. BTW: you can make use of this fact:\[a^2\gt a^2-1\]\[\therefore a^2\gt (a+1)(a-1)\]In this case a=38 so you know that:\[38^2\gt (38+1)(38-1)\]
anonymous
  • anonymous
how did you just decide a = 38? @asnaseer
asnaseer
  • asnaseer
in your last step you had:\[38^2?39\times37\]
asnaseer
  • asnaseer
so can conclude ? is ">" without actually evaluating \(38^2\)
anonymous
  • anonymous
but 38^2 is "a" but 39 x 37 is not = (a+1)(a-1) so how can you compare that way?
asnaseer
  • asnaseer
no, a=38, remember the identity I showed was:\[a^2\gt (a+1)(a-1)\]so, if a=38 we get:\[38^2\gt(38+1)(38-1)\]\[\therefore 38^2\gt39\times37\]
anonymous
  • anonymous
wow!! i see it now. where did you know this. i never came across this before?
asnaseer
  • asnaseer
nor did I - I just suddenly "spotted" the pattern - that is one of the beauty of maths, you never know what subtleties you will discover. :)
anonymous
  • anonymous
thanks...
asnaseer
  • asnaseer
yw :)

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