if x^2-2011x+1=0 is true,then waht is the value of x-2010x+2011/(x^2+1)?

quinn
Sep 25, 2014

#1**+8 **

$$\\If\;\; x^2-2011x+1=0 $ is true, then what is the value of $\\\\

x-2010x+\frac{2011}{(x^2+1)}\;\;?\\\\

$first$\\

x^2-2011x+1=0\\

x^2+1=2011x\\\\

$now$\\\\\

x-2010x+\frac{2011}{(x^2+1)}\\\\

=x-2010x+\frac{2011}{(2011x)}\\\\

=x-2010x+\frac{1}{x}\\\\

=-2009x+\frac{1}{x}\\\\$$

Maybe you needed more brackets in your question? I think that you probably did.

Melody
Sep 25, 2014

#5**+8 **

$$\\If\;\; x^2-2011x+1=0 $ is true, then what is the value of $\\\\

x-2010x+\frac{2011}{(x^2+1)}\;\;?\\\\

$first$\\

x^2-2011x+1=0\\

x^2+1=2011x\\\\

$now$\\\\\

x^2-2010x+\frac{2011}{(x^2+1)}\\\\

=x^2-2010x+\frac{2011}{(2011x)}\\\\

=x^2-2010x+\frac{1}{x}\\\\

=(x^2+1)-1-2010x+\frac{1}{x}\\\\

=2011x-1-2010x+\frac{1}{x}\\\\

=x-1+\frac{1}{x}\\\\$$

$$\\=\frac{x^2-x+1}{x}\\\\

=\frac{1011x-x}{x}\\\\

=\frac{1010x}{x}\\\\

=1010$$

I think I might have stuffed up somewhere. Plus I have a reputation for doing things the LONG way.

I'm going to look over it some more.

Melody
Sep 25, 2014

#9**+13 **

Best Answer

x^{2} = 2011x - 1 → x^{2} + 1 = 2011x

So

x^2-2010x+【2011/(x^2+1)] =

[2011x - 1 ] - 2010x + 2011/(2011x) =

x - 1 + 1/x =

[x^{2} - 1x + 1] / x =

[x^{2} + 1 - x ] /x =

[2011x - x] / x =

2010x / x =

2010 (with the restriction that x ≠ 0)

CPhill
Sep 25, 2014

#10**+8 **

I thought about the restriction Chris but I didn't put it there because the first statement would not be true if x=0.

But like always - yours is shorter.

I started mine from the previous result where the question was wrong.

That is my excuse and I am sticking with it! LOL

Melody
Sep 25, 2014

#12**+8 **

(x^2-2011x+1)/x=0/x

x-2011+1/x=0

x+1/x=2011

x-1+1/x=2011-1=2010

Good job.Is it funny or not to do problem like this?if yes, i will put a problem everyday to the forum

quinn
Sep 25, 2014

#13**+8 **

It * IS* fun, quinn.......just don't make them too difficult......otherwise, it makes us look bad........

P.S ....I even had to "cheat" off Melody a bit to get that problem correct!!!! (Shhh!!!...don't tell her)

CPhill
Sep 25, 2014

#14**+8 **

Yes I had fun with this problem quinn.

But it is better if you get the question right in the first place. I liked the Extra brackets that you put in second time round, then we know we have not misinterpreted.

I have included this problem in the next "End of Day Wrap" that i write. I invite you to read my wrap. I write one most nights.

Here is the address of yesterday's one. They are kept in with the sticky topics. :)

http://web2.0calc.com/questions/all-questions-answered-to-this-point#r294

(I don't usually include that many questions but I had not written one for a few days)

We really DID LIKE this question Quinn, you are welcome here any time

Melody
Sep 25, 2014