Dave Posted August 21, 2003 Posted August 21, 2003 I'm bored, so here's a nice piece of maths: Simplify sqrt(2 + sqrt(3)) - sqrt(2- sqrt(3)). A few of you may have already seen this - if so, don't spoil it for the others. It's not supposed to be hard, but it's fairly nice.
YT2095 Posted August 22, 2003 Posted August 22, 2003 never seen it, but at 1`st glance looks a bit self defeating, and so I`de guess at Zero? I`ll check when I get more time though
Dave Posted August 22, 2003 Author Posted August 22, 2003 Obviously not zero. first term clearly != second term. Hint: put it in your calculator, then try and simplify. (yes, it's cheating )
YT2095 Posted August 22, 2003 Posted August 22, 2003 surely if the 1`st term = X and the second = X also then a subtraction of X-X=0 regardless of what X might equal after the 1`st term? (still haven`t grabbed ma calc yet tho )
YT2095 Posted August 22, 2003 Posted August 22, 2003 Oooops MY bad! I didn`t see the `-` in the second term were used + in the 1`st.... term 1 and 2 looked the same.... forgive my oversight
quack Posted August 22, 2003 Posted August 22, 2003 I used my calculator and got square root of 2 (actually 1.4 etc.). What is the reasoning behind it?
Dave Posted August 23, 2003 Author Posted August 23, 2003 I'll tell you all a bit later on if you haven't got it by then
Dave Posted August 24, 2003 Author Posted August 24, 2003 Okay, I can't be bothered waiting for someone to reply, so here's the answer. Let x = sqrt(2 + sqrt(3)) - sqrt(2- sqrt(3)) This means x^2 = 4 - 2*sqrt((2+sqrt(3))(2-sqrt(3)) = 4 - 2 * sqrt(1) = 2 Therefore x = sqrt(2) There we go. Ta da etc.
cHIs- Posted August 24, 2003 Posted August 24, 2003 dave said in post #1 :I'm bored, so here's a nice piece of maths: Simplify sqrt(2 + sqrt(3)) - sqrt(2- sqrt(3)). A few of you may have already seen this - if so, don't spoil it for the others. It's not supposed to be hard, but it's fairly nice. WTF!?
Dave Posted August 24, 2003 Author Posted August 24, 2003 ? There ain't nothin hard about it, it's just seeing a method.
Dave Posted August 24, 2003 Author Posted August 24, 2003 read up some more on algebra and it'll become a bit more apparent.
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