## Where did the other root go?

### 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?

$\Rightarrow e^{x}+e^{-x}=2$

$\Rightarrow e^{2x}-2e^{x}+1=0$

$\Rightarrow e^{x}=1\pm \sqrt{2}$

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?

### 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.

### 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 $n=0$.

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.

- Zero.

- Why?

- Because all the poles are in Eastern Europe.

Revive the IMO marathon.

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

Sorry, i made a mistake.. Correcting it below:

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

$\Rightarrow e^{x}-e^{-x}=2$

$\Rightarrow e^{2x}-2e^{x}-1=0$

$\Rightarrow e^{x}=1\pm \sqrt{2}$

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

$\Rightarrow e^{x}-e^{-x}=2$

$\Rightarrow e^{2x}-2e^{x}-1=0$

$\Rightarrow e^{x}=1\pm \sqrt{2}$

**The rest part follows now...**### 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.

- Zero.

- Why?

- Because all the poles are in Eastern Europe.

Revive the IMO marathon.

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

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

### 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.

- Zero.

- Why?

- Because all the poles are in Eastern Europe.

Revive the IMO marathon.

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

Sorry didn't see it