Q2: Solve for x, y, z

√x + √y + √z = 4
x^2 + y^2 + z^2 = 40.125
e^(xyz) = 4.771

Question

Q2: Solve for x, y, z 

√x + √y + √z = 4 
x^2 + y^2 + z^2 = 40.125 
e^(xyz) = 4.771

aric200 · Answer

@satellite73

anonymous · Answer

there should be a limit on how sadistic maths teachers should get :>

agent0smith · Answer

take logs of both sides of the last equation$$\large xyz = \ln 4.771 $$ i'm sure you can solve using substitution with the first two equations.

aric200 · Answer

I honestly have no idea what the question is even talking about I just looked up the hardest algebra question I could and wanted someone to solve it :O

agent0smith · Answer

Oh. Then that gives me even less reason to try to solve it for you.

aric200 · Answer

I suppose you are right..

agent0smith · Answer

It looks like a lot of work to solve, not in a fun challenging way, just a tedious lots-of-algebra kinda way.

aric200 · Answer

Oh here are the answers But im going to ask real questions if you can maybe assist me with them?
x=1, y=0.25, z=6.25

agent0smith · Answer

lol i didn't need the answers :P wolfram alpha could've given those. And sure.

aric200 · Answer

Wolfram good site good site. haha

agent0smith · Answer

lol good thing i didn't try to solve it: http://www.wolframalpha.com/input/?i=%E2%88%9Ax+%2B+%E2%88%9Ay+%2B+%E2%88%9Az+%3D+4%2C++x%5E2+%2B+y%5E2+%2B+z%5E2+%3D+40.125%2C++e%5E%28xyz%29+%3D+4.771+ wolfram even gave up.