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How can I solve this for $x$?

$xe^x=-2/a$ with $(a \in \mathbb{R_0^+})$

$a$ can be any strict positive real number.

I need this because I'm searching for the root of a function to sketch a graph.

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    Mats, the reals are a subset of the complex numbers so there's no issue. You can see what $W$ looks like on $\mathbb{R}$ on the graphic at the top of the linked Wikipedia page. The equation $xe^x = b$ **only** has real solution(s) if $b\ge -e^{-1}$. Also, nobody takes the chat warning seriously. :)2011-08-24

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I suspect anon's comment is what your exam paper was looking for: for $a$ sufficiently small, $ax e^x + 2 > 0$ always, so for a sketch of the graph you have something smooth with certain asymptotic behaviour.

For $a = 2e$, there is exactly one root, and you should show that you get blow-up of the same sign when you approach the singular point from the left and when you approach the singular point from the right. (Informally: $\lim_{x\to -1^+} (axe^x + 2)^{-1} = \lim_{x\to -1^-} (axe^x + 2)^{-1} = +\infty$)

For $a > 2e$, there are two roots, and at those two singularities, the left limit and the right limit have opposite signs, so you need to show that on the graph. In other words, I suspect the exam paper was asking for a qualitative depiction of the graph of the function, and not for a strictly quantitative depiction.