# X^2+x + lnx

1. Dec 10, 2012

### Bonaparte

1. The problem statement, all variables and given/known data

So I have happily exploring function when I got to the equation 0 = 1/2(e^(2x))-(e+1)(e^x) +ex.

2. Relevant equations

Well, I guess the quadratic formula can help, although I can't seem to get to a situation where I can use it.

3. The attempt at a solution
I played around until I got to t^2-(2e+2)t+2eln(t) where t=e^x. But I seem to always get stuck with that ex, which does not let me factor the e^x.

2. Dec 10, 2012

### Michael Redei

Re: x^2+x+lnx

What's your problem? What are you trying to do with this function?

Sorry, now I see that you're trying to solve an equation...

3. Dec 10, 2012

### Michael Redei

Re: x^2+x+lnx

All I can offer is an approximation, x = 1.75566566912961...

4. Dec 10, 2012

### Bonaparte

Re: x^2+x+lnx

But how would you find it?

5. Dec 10, 2012

### Ray Vickson

Re: x^2+x+lnx

You need to solve the equation f(x) = x^2 + x + ln(x) = 0 numerically. There is a vast literature on this, but basically, you first need to isolate a region in which the root lies, then try to narrow it down. If you can, you should plot the graph y = f(x) first, to see roughly where the roots of f(x) are located. Then there are numerous "correction" methods available to get better accuracy; just Google 'root finding' to see lots of relevant methods. For example, look at http://www.efunda.com/math/num_rootfinding/num_rootfinding.cfm

6. Dec 10, 2012

### Michael Redei

7. Dec 10, 2012

### Ray Vickson

Re: x^2+x+lnx

That's what I would do too, but I would use Maple. The solution of the equation $0 = x^2 + x + \ln(x)$ is approximately .4858388639605664330809376128591963662449, and is nowhere near the value 1.755... that you wrote. If you plot f(x) = x^2 +x +ln(x) on [0.001,6] you will see there is just one root, and it is near x = 1/2. Were you working with the other function
$$f(x) = \frac{1}{2} e^{2x} - (e+1)e^x + ex ?$$ That does, indeed, have a root near 1.75.

8. Dec 10, 2012

### Michael Redei

Re: x^2+x+lnx

When I posted the approximation 1.75..., "the other function" was what seemed the main goal of this thread. "Were you working" seems a little exaggerated though, since I did no more than type a function definition into one text box and copy a number from another.