tomiko
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
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tomiko
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attched is my working
jasonxx
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seems correct
tomiko
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thanks a lot!
jasonxx
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(:
asnaseer
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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:\[38^2\gt (38+1)(38-1)\]
tomiko
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how did you just decide a = 38? @asnaseer
asnaseer
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in your last step you had:\[38^2?39\times37\]
asnaseer
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so can conclude ? is ">" without actually evaluating \(38^2\)
tomiko
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but 38^2 is "a" but 39 x 37 is not = (a+1)(a-1) so how can you compare that way?
asnaseer
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no, a=38, remember the identity I showed was:\[a^2\gt (a+1)(a-1)\]so, if a=38 we get:\[38^2\gt(38+1)(38-1)\]\[\therefore 38^2\gt39\times37\]
tomiko
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wow!! i see it now. where did you know this.
i never came across this before?
asnaseer
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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. :)
tomiko
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thanks...
asnaseer
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yw :)