Find roots for an equation with quadratic, linear and log terms?

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I'm wondering if there exists a closed-form or analytic expression for the roots of an equation of the form

$ax^2 + bx + c\log x=0.$

considering the natural $\log$. Wolfram alpha is leading me to expressions involving the Lambert W (product log) function when I include either the quadratic term or the linear term (but not both) and analytic approximations when I supply real values for the coefficients.

This is OK, but does a more general solution exist in terms of the coefficients?

thanks!

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Any equation which can write $$A+B x+C \log(D+Ex)=0$$ has a solution which can be expressed in terms of Lambert function. $$x=\frac{C }{B}W(F)-\frac{D}{E} $$ where $$F=\frac{B }{C E} e^{\frac{B D-A E}{C E}}$$This is why Wolfram Alpha gave you solutions for $$ax^2 + bx + c\log x=0$$ in the case where $a=0$ or $b=0$.

However, for the general case you are considering $(a \neq 0,b\neq 0)$, I do not think that exists any analytical solution and only numerical methods (such as Newton) would solve the problem and this should be quite easy.

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The equation $ax^2 + bx + c\log x=0$ can be rewritten as (supposing a not null): $$\frac{a}{c}(x+\frac{b}{2a })^2-\frac{b^2}{4a c}+ \log(x)=0$$ Taking the exponential of both members: $$e^{-\frac{a}{c}(x+\frac{b}{2a })^2+\frac{b^2}{4a c}}=x$$ Remembering Lagrange series: $$x=\sum_{n=1} \frac{D^{n-1} \left. e^{-n\frac{a}{c}(x+\frac{b}{2a })^2 + \frac{nb^2}{4a c}}\right|_{x=0}}{n!}$$

$$x=\sum_{n=1} e^{n\frac{b^2}{4a c}} \frac{D^{n-1} \left. e^{-n\frac{a}{c}(x+\frac{b}{2a })^2}\right|_{x=0}}{n!}$$ Supposing $a/c > 0$ and remembering Rodrigues formula of Hermite polynomials: $$H_n(x)=(-1)^n e^{x^2}\frac{d^n}{dx^n}e^{-x^2}$$ we have: $$x=\sum_{n=1}^{\infty} \frac{ \left( -\sqrt\frac{na}{c} \right)^{n-1} H_{n-1}\left( \sqrt\frac{na}{c} \frac{b}{2a}\right) }{n!}$$

References

"Generalization of Lambert $W$ function, Bessel polynomials and transcendental equations", Giorgio Mugnaini, http://arxiv.org/pdf/1501.00138