## Where did the other root go?

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Rabeeb
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Joined: Tue Dec 14, 2010 7:52 pm
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### Where did the other root go?

Let's solve the equation: $\frac{e^{x}+e^{-x}}{2}=1$
$\Rightarrow e^{x}+e^{-x}=2$
$\Rightarrow e^{2x}-2e^{x}+1=0$
$\Rightarrow e^{x}=1\pm \sqrt{2}$
But $e^{x }$ cannot be negative, so $x= ln (1+ \sqrt {2}) \approx 0.881$
But from the equation, it is clearly observable that $-0.881$ is an approx root too, which we did not obtain in the normal process. Where did it go?

SMMamun
Posts: 57
Joined: Thu Jan 20, 2011 6:57 pm

### Re: Where did the other root go?

It didn't go anywhere. You have made an error in the root extraction formula for the quadratic form of equations. Please check again. Nirjhor
Posts: 136
Joined: Thu Aug 29, 2013 11:21 pm
Location: Varies.

### Re: Where did the other root go?

Notice that $e^{2x}-2e^x+1=\left(e^x-1\right)^2=0\Rightarrow e^x=1$ so $x=0$ is the only solution.

Another way to observe is that the left side is the definition of hyperbolic cosine. So $\cosh x=1$ gives $x=2\pi i n$ for all $n\in\mathbb{Z}$. Only real solution is $x=0$ for $n=0$.
- What is the value of the contour integral around Western Europe?

- Zero.

- Why?

- Because all the poles are in Eastern Europe.

Revive the IMO marathon.

Rabeeb
Posts: 25
Joined: Tue Dec 14, 2010 7:52 pm
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### Re: Where did the other root go?

Sorry, i made a mistake.. Correcting it below:
$\frac{e^{x}-e^{-x}}{2}=1$
$\Rightarrow e^{x}-e^{-x}=2$
$\Rightarrow e^{2x}-2e^{x}-1=0$
$\Rightarrow e^{x}=1\pm \sqrt{2}$
The rest part follows now...

Nirjhor
Posts: 136
Joined: Thu Aug 29, 2013 11:21 pm
Location: Varies.

### Re: Where did the other root go?

Then how is $-\ln\left(1+\sqrt 2\right)$ another root?
- What is the value of the contour integral around Western Europe?

- Zero.

- Why?

- Because all the poles are in Eastern Europe.

Revive the IMO marathon.

Rabeeb
Posts: 25
Joined: Tue Dec 14, 2010 7:52 pm
Contact:

### Re: Where did the other root go?

Nirjhor wrote:Then how is $-\ln\left(1+\sqrt 2\right)$ another root?
Replace x by -x in the equation, and -x by x. Do u see any change?

Nirjhor
Posts: 136
Joined: Thu Aug 29, 2013 11:21 pm
Location: Varies.

### Re: Where did the other root go?

Yes I do. Then $\dfrac{e^x-e^{-x}}{2}$ changes to $\dfrac{e^{-x}-e^x}{2}$. They're obviously not equal. They'd be if the sign was 'plus'. But we've already dealt with that case.
- What is the value of the contour integral around Western Europe?

- Zero.

- Why?

- Because all the poles are in Eastern Europe.

Revive the IMO marathon.

Rabeeb
Posts: 25
Joined: Tue Dec 14, 2010 7:52 pm
Sorry didn't see it 