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Polynomial equation

Posted: Thu Mar 03, 2011 10:28 am
by jagdish
Calculate value of $x$ in $(x+1)(x+2)(x+3)(x+4)=120x^2$

Re: Polynomial equation

Posted: Thu Mar 17, 2011 1:59 pm
by nfstiham
probably 1

Re: Polynomial equation

Posted: Wed Mar 23, 2011 6:33 pm
by jagdish
Now How can I Calculate other Roots.

Re: Polynomial equation

Posted: Mon Mar 28, 2011 7:13 pm
by *Mahi*
I think with $x$ excessing a limited value,$(x+1)(x+2)(x+3)(x+4)>120x^2$ so we will find the limit of $x$ and then finding the $x$ 's is not at all hard.(for the first part ,you have to expand the L.H.S. and do some argument like, for $x>a,x^3>ax^2$)

Re: Polynomial equation

Posted: Sat Apr 02, 2011 9:58 pm
by Masum
jagdish wrote:Calculate value of $x$ in $(x+1)(x+2)(x+3)(x+4)=120x^2$
But you didn't say what is $x$?
$x\in \mathbb{N}$ or else?Since posted in Algebra fora,$x$ can be anything.
Whatever,you may limit the values of $x$ using inequalities as the previous post says.

Re: Polynomial equation

Posted: Thu Feb 28, 2013 2:11 am
by zadid xcalibured
$(x^2+5x+4)(x^2+5x+6)=120x^2$
$\Longrightarrow (x^2+5x+5)^2-1=120x^2$
$\Longrightarrow (x^2+5x+5)^2-121x^2=1-x^2$
$\Longrightarrow (x^2+16x+5)(x-5)(x-1)=(x+1)(1-x)$
so $1$ is a root.Eliminating this root their remains a cubic equation which can be solved with a routine analogy.