Math Is Fun Forum

tony123

Member

Registered: 2007-08-03

Posts: 230

without a calculator

without a calculator prove

ZHero

Real Member

Registered: 2008-06-08

Posts: 1,889

Re: without a calculator

Consider
2 + rt3 = 1/2 + 3/2 + 2 1/2 rt3
= [rt(1/2)]^2 + [rt(3/2]^2 + 2 rt(3/2) rt(1/2)
---> 2 + rt3 = [rt(1/2) + rt(3/2)]^2
therefore..
rt(2 + rt3) = rt(1/2) + rt(3/2)
similarly..
rt(4 - rt7) = rt(7/2) - rt(1/2)
rt(5 + rt21) = rt(7/2) + rt(3/2)

substitute the values in LHS & RHS n its proved!!
Kindly don't mind my not so neat method of writing but actually i'm using my mobile to surf n don't really know how to write codes..

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If two or more thoughts intersect, there has to be a point!

ZHero

Real Member

Registered: 2008-06-08

Posts: 1,889

Re: without a calculator

Another way to do it would be..
Let..
rt(2 + rt3) = x + rty
2 + rt3 = x^2 + y + 2x(rty)
x^2 + y = 2 ...(1)
&
2x(rty) = rt3
---> y = 3/(4x^2) ...(2)
putting in (1)
x^2 + 3/(4x^2) = 2 which is quadratic in x^2.
Considering only positive roots, we get..
x = rt(1/2) & rt(3/2)
which gives
y = rt(3/2) & rt(1/2)
etc. etc.

Thus simplify all the surds & you'll get the answer!!!
Regards...

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If two or more thoughts intersect, there has to be a point!