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Mathematics 14 Online
OpenStudy (anonymous):

How about this... \frac {3}{3+\frac {3}{3+\frac {3}{3+\frac {3}{...}}}}

OpenStudy (anonymous):

\[\frac {3}{3+\frac {3}{3+\frac {3}{3+\frac {3}{...}}}}\]

OpenStudy (anonymous):

x^2 + 3x = 3

OpenStudy (lgbasallote):

let x equal that 3/3+x = x 3 = 3x + x^2 x^2 + 3x - 3 =0 -3 +/- sqrt (9 - 4(1)(-3)/2(1) -3 +/- sqrt (9 + 12) /2 -3 +/- sqrt 21/2 x= 0.7913

OpenStudy (lgbasallote):

there seems to be something wrong with my quadratic formula... x^2 + 3x = 3 x(x+3) = 3 x = 3 x = 0 ???

OpenStudy (anonymous):

if x=3, isnt the rhs=18? @Igbasallote

OpenStudy (anonymous):

and if x=0, rhs=0

OpenStudy (lgbasallote):

oh yeah...so my quadratic formula is right...my factor was the one that's wrong =))))))

OpenStudy (apoorvk):

|dw:1332670830662:dw| simplify and enjoy!!!

OpenStudy (callisto):

Sorry to ask, how can you guys get to this step: 3/3+x = x ?

OpenStudy (apoorvk):

callisto, i hope you can see that first the whole expression's value to be 'x'. then the part in denominator after "3+.." that is just 'x' as well too right?? because the whole equation continues to divided infinite times!!!

OpenStudy (callisto):

you let it be x , so it is x...

OpenStudy (callisto):

Nah.. it seems i understand a little..

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