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dave
08-21-2003, 01:00 PM
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
08-22-2003, 09:53 AM
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
08-22-2003, 10:13 AM
Obviously not zero. first term clearly != second term.

Hint: put it in your calculator, then try and simplify. (yes, it's cheating :-p)

YT2095
08-22-2003, 10:30 AM
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
08-22-2003, 10:32 AM
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

dave
08-22-2003, 01:28 PM
Yup, they're two different terms.

quack
08-22-2003, 04:14 PM
I used my calculator and got square root of 2 (actually 1.4 etc.). What is the reasoning behind it?

dave
08-23-2003, 04:06 AM
I'll tell you all a bit later on if you haven't got it by then :)

dave
08-23-2003, 05:09 PM
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-
08-23-2003, 07:17 PM
dave said in post #1 (http://www.scienceforums.net/forums/showthread.php?s=&postid=19814#post19814):
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!? :D

dave
08-24-2003, 02:31 AM
?

There ain't nothin hard about it, it's just seeing a method.

cHIs-
08-24-2003, 08:59 AM
its hard when your me :P

dave
08-24-2003, 03:46 PM
read up some more on algebra and it'll become a bit more apparent.