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Callisto
Tutor question #1 Simplify the following: \[\frac{1}{\sqrt2 + \sqrt3 + \sqrt5}\] (conditions: pen, paper, no calculator, done in 2 minutes) *
I got \[ \frac{2\sqrt3 + 3\sqrt2 - \sqrt{30}}{12}\] Was that right?
HUNDRED PERCENT!!!!!!!!!!!!!
But it looks so messy...
i got something different: isn't it you cant have radicals under a fraction so you multiply by the conjugate?
That's what rationalise means
Not messy at all, this is the only way to go about it.
yes but if you multiply 1 by the conjugate pair you get the conjugate pair.. your numerator seems wrong..
otherwise can someone please show me a step by step to how you got that answer? and i'll show you what i did :)
luckily, I haven't deleted that.. \[(\sqrt2 + \sqrt3 + \sqrt5)(\sqrt2 + \sqrt3 - \sqrt5)\]\[ = (\sqrt2 + \sqrt3)^2-5\]\[ = 2 + 3 + 2\sqrt6-5\]\[=2\sqrt6\] \[\frac{1}{\sqrt2 + \sqrt3 + \sqrt5}\]\[ = \frac{\sqrt2 + \sqrt3 - \sqrt5}{(\sqrt2 + \sqrt3 + \sqrt5)(\sqrt2 + \sqrt3 - \sqrt5)}\]\[ = \frac{\sqrt2 + \sqrt3 - \sqrt5}{2\sqrt6}\]\[ = \frac{2\sqrt3 + 3\sqrt2 - \sqrt{30}}{12}\]
\[1/(\sqrt{2}+\sqrt{3}+\sqrt{5})=1/(a+\sqrt{5})=(a-sqrt{5})/(a^{2}-5)\] \[(\sqrt{2}+\sqrt{3}-\sqrt{5})/(5-5+\sqrt{6})\] \[\sqrt{2}+\sqrt{3}-\sqrt{5}/2\sqrt{6}=2.\sqrt{3}+2.\sqrt{2}-\sqrt{30}/12\]
I ended at the second to last step... where did you get the 2 in front of the radicals if you are multiplying by 1?
see the first part of it..
5-5+2sqrt(6)*
@RaphaelFilgueiras The second term of the numerator is 3\sqrt{2} ?! \sqrt{18} = \sqrt{9 times 2} = 3\sqrt2 ...
okay, so the you rationalized the second to last line because of the radical at the bottom?
Of course! That's 'rationalise'
okay sorry, then you are right.. I just wanted to know where I went wrong...
no in my solution i put (5-5+sqrt(5))(WRONG) instead(5-5+2sqrt(5))
Sorry, but I don't know what you're doing.. No offense.
Do you mind re-type it again?
i did same as you,but call(a=sqrt(2)+sqrt(3))
Okay, and the answer is the same?
Okay, thanks everyone!!!