Polynomial equation

For discussing Olympiad Level Algebra (and Inequality) problems
jagdish
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Polynomial equation

Unread post by jagdish » Thu Mar 03, 2011 10:28 am

Calculate value of $x$ in $(x+1)(x+2)(x+3)(x+4)=120x^2$
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nfstiham
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Re: Polynomial equation

Unread post by nfstiham » Thu Mar 17, 2011 1:59 pm

probably 1

jagdish
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Re: Polynomial equation

Unread post by jagdish » Wed Mar 23, 2011 6:33 pm

Now How can I Calculate other Roots.
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*Mahi*
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Re: Polynomial equation

Unread post by *Mahi* » Mon Mar 28, 2011 7:13 pm

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 $x>a,x^3>ax^2$)
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Masum
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Re: Polynomial equation

Unread post by Masum » Sat Apr 02, 2011 9:58 pm

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.
One one thing is neutral in the universe, that is $0$.

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zadid xcalibured
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Re: Polynomial equation

Unread post by zadid xcalibured » Thu Feb 28, 2013 2:11 am

$(x^2+5x+4)(x^2+5x+6)=120x^2$
$\Longrightarrow (x^2+5x+5)^2-1=120x^2$
$\Longrightarrow (x^2+5x+5)^2-121x^2=1-x^2$
$\Longrightarrow (x^2+16x+5)(x-5)(x-1)=(x+1)(1-x)$
so $1$ is a root.Eliminating this root their remains a cubic equation which can be solved with a routine analogy.

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