Solving a quadratic equation involving e

Question

I am trying to solve the following equation for x. Is there an algebraic way to solve this equation

$$x^2 + x + e^{\frac{x}{x+3}}= 10$$

My objective is find he value of $x$, given a set of other parameters

Answer 1

Transcendental equations do not show analytical solutions (notice that this is already the case of $\cos(x)=x$) and numerical methods are required.

Probably the simplest would be Newton method which starting from a "reasonable" guess $x_0$ will update it according to $$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}$$ In your case $$f(x)=x^2 + x + e^{\frac{x}{x+3}}- 10$$ $$f'(x)=2 x+\frac{3 e^{\frac{x}{x+3}}}{(x+3)^2}+1$$ Looking at the plot, a solution seems to be close to, say, $x_0=2$. So applying the method, you would get as successive iterates $$\left( \begin{array}{cc} n & x_n \\ 0 & 2.0000000000000000000 \\ 1 & 2.4842954466725154721 \\ 2 & 2.4469077855075530182 \\ 3 & 2.4466816478021901217 \\ 4 & 2.4466816395302785131 \\ 5 & 2.4466816395302785021 \end{array} \right)$$

Just for your curiosity, you will learn sooner or later that, better than with Taylor series, a function can be approximated with the so-called Pade approximants. The simplest, built around $x=a$ would be given by $$f(x)\approx \frac{f(a)+\frac{ \left(2 f'(a)^2-f(a) f''(a)\right)}{2 f'(a)}(x-a)}{1-\frac{ f''(a)}{2 f'(a)}(x-a)}$$ Ccancelling the numerator, we should have an approximation of the solution $$x=a+\frac{2 f(a) f'(a)}{f(a) f''(a)-2 f'(a)^2}$$ Applied to your problem, using $a=2$, we should get $x=2.44383$ which is not too bad (I hope).

Answer 2

An algebraic way does not exist because your equation is transcendent.

I think the following equation is nice already. $$x^2+x+2^{x^2-4x+6}=10$$