Where does this plane curve intersect the x-axis?

Question

In a problem in electrostatics, I came across this plane curve: $$ 4\frac{x+2}{\sqrt{ (x+2)^{2}+y^{2} }}-\frac{x+1}{\sqrt{ (x+1)^{2}+y^{2} }}=3, $$ and by plotting the curve I found that it crossed the $x$-axis at $x=-2$ and $x=0$. But how (if possible) can these solutions be obtained analytically?

For example, substituting $x=-2$ gives the relation $$ \frac{1}{\sqrt{1+y^2}}=3 $$ which only has pure-imaginary solutions for $y$, while we would naively expect $y=0$ to be a solution. Clearly, the curve is not well-behaved, but are there better approaches than graphing to determine its behavior?

(For some additional context on things I'd like to be able to infer from this curve, the problem asks to find the angle relative to the $x$-axis of this curve at $x=-2$, which should be 60 degrees. Using the curve directly is not the only approach to find the angle but it is the motivating problem.)

Answer 1

Explore the neighborhood of $(x,y)=(-2,0)$, set $x=-2+\varepsilon c$, $y=εs$, where $ε\gtrapprox 0$, $c^2+s^2=1$. Then the equation reads $$ 4c-\frac{-1+εc}{\sqrt{1-2εc+ε^2}}=3 $$ So $c=\frac12+O(ε^2)$, which indeed gives an angle of $60^°$ to the $x$-axis.

Answer 2

Consider $$ 4\frac{x+2}{\sqrt{ (x+2)^{2}+y^{2} }}-\frac{x+1}{\sqrt{ (x+1)^{2}+y^{2} }}=3. $$ Set $y=0$ to get $$ 4\frac{x+2}{\sqrt{ (x+2)^{2}}}-\frac{x+1}{\sqrt{ (x+1)^{2}}}=3 \\ 4\frac{x+2}{|x+2|}-\frac{x+1}{|x+1|}=3 \\ 4\operatorname{sgn}(x+2)-\operatorname{sgn}(x+1)=3 $$ Consider the possibilities for $x$: \begin{align} \text{if } x>-1:&\qquad 4-1=3, \qquad\text{true} \\ \text{if } x=-1:&\qquad 4-0=3, \qquad\text{false} \\ \text{if } -2<x<-1:&\qquad 4-1=3, \qquad\text{false} \\ \text{if } x=-2:&\qquad 0-1=3, \qquad\text{false} \\ \text{if } x<-2:&\qquad -4-1=3, \qquad\text{false} \end{align}

So, for solutions I get: all real numbers $x$ with $x>-1$.