if x+sqrt(x)=1 what is the value of x^2+2x.sqrt(x)
\[x+\sqrt{x}=1\] \[\sqrt{x}=1-x\]x^2-3x+1=0 \[x=1-2x+x^2\] \[x^2-3x+1=0\] \[x=\frac{3-\sqrt{5}}{2}\] and \[1=(x+\sqrt{x})^2=x^2+2x\sqrt{x}+x\]so \[x^2+2\sqrt{x}=1-\frac{3-\sqrt{5}}{2}=\frac{2-3+\sqrt{5}}{2}=\frac{-1+\sqrt{5}}{2}\]
where did you get x^2-3x+1=0 from? Oo
\[\sqrt{x}=1-x\] square both sides get \[x=(1-x)^2=1-2x+x^2\] set equal to zero to solve the quadratic get \[x^2-3x+1=0\] unless i goofed up
no, you're right, i just didn't pay attention :P
you really in brazil?!
haha yeah lol
wow.
near the ocean?
10 min :P
nice
so if i sometimes say something wrong in my questions, it's becaus i have to translate them all :)
from portuguese?
yup. all of them :)
i am impressed
be
why not? i can barely speak one language, and portuguese is way different than english
yeah but i lived in england for a few years when i was little, so i still remember a few things. The hard part is to know scientific words, but MIT open course helps me with that lol
i bet brazil in nicer than england
no way, i'm doing everything to go back. where are you from?
philadelphia
going back when you graduate? or sooner?
sooner, my dad works in reasearch and he knows a lot of people at some universities in england so i might try to get a transfer or something when he goes there to do his Post-doc
good luck! hope you get there soon.
thank you :)
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