Congratulations @sauravshakya and @ghazi for Golden..
OpenStudy (anonymous):
Thanx @waterineyes
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OpenStudy (anonymous):
What do you mean by x common?
OpenStudy (lgbasallote):
congratulations @lgbasallote for beating @saifoo.khan finally
OpenStudy (ghazi):
@waterineyes thanks....
OpenStudy (anonymous):
X^3-x^2-2x+1=0
x(x^2-x-2+1/x)=0
Parth (parthkohli):
@lgbasallote that's epic, really!
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OpenStudy (anonymous):
I just took x common from X^3-x^2-2x+1=0
OpenStudy (anonymous):
Welcome to both..
OpenStudy (ghazi):
\[x^3-x^2-2x+1=0....x(x^2-x-2+\frac{ 1 }{ x })\]
OpenStudy (ghazi):
@waterineyes got to know about you...
OpenStudy (anonymous):
@Ausam ????
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OpenStudy (anonymous):
What @ghazi ??
OpenStudy (ghazi):
you're a girl.. :)
OpenStudy (anonymous):
Sorry. Press by mistake
OpenStudy (anonymous):
Having any doubt ??
OpenStudy (anonymous):
Yeah
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OpenStudy (anonymous):
Ha ha ha ha...
OpenStudy (ghazi):
before lookin at your profile...i was in doubt..but first solve this doubt
OpenStudy (anonymous):
I still don't get how to do it clearly
OpenStudy (anonymous):
I am asking to @ghazi
Anyways what is your doubt??? @asdfasdfasdfasdfasdf1
OpenStudy (ghazi):
@Ausam take out x from equation
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OpenStudy (anonymous):
X^3-x^2-2x+1=0
x(x^2-x-2+1/x)=0
x^2-x-2+1/x=0
OpenStudy (anonymous):
First clear the doubt of @asdfasdfasdfasdfasdf1 then we will talk..
OpenStudy (ghazi):
okay
OpenStudy (anonymous):
Now, arrange the left hand side as the question............ ANd took the rest to right hand side
OpenStudy (ghazi):
@Ausam divide whole equation by X
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OpenStudy (anonymous):
Thanks. Thatd be great
OpenStudy (anonymous):
Oh really sorry gotta go now I will clear your doubt next time I will come online..
Sorry for that..
I am in college computer lab and now it is time to close the lab..
So you have to wait till tomorrow.. @ghazi
OpenStudy (ghazi):
it is cool
OpenStudy (anonymous):
So I factorize using x nd then arrange it like the equation given?
OpenStudy (ghazi):
\[\frac{ x^3-x^2-2x+1 }{ x }=\frac{ x^3 }{ x }-\frac{ x^2 }{ x }-\frac{ 2x }{ x }+\frac{ x }{ x }\] @Ausam clear?
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