Problem 1.2 (E&M Purcell & Morin) Zero force from a triangle, multiple solutions?

Problem 1.2 of the E&M text of Purcell & Morin reads:

"Two positive ions and one negative ion are fixed at the vertices of an equilateral triangle. Where can a fourth ion be placed, along the symmetry axis of the setup, so that the force on it will be zero? Is there more than one such place? You will need to solve something numerically."

The solution then says: "Let the sides of the triangle be 2 units long. Consider a point $P$ that lies a distance
$y$ (so $y$ is defined to be a positive number) beyond the side containing the
two positive ions. $P$ is a distance $y + sqrt{3}$ from the negative ion and $sqrt{1+y^2}$ from each of the positive ions. If the electric field equals zero at $P$, then the upward field due to the negative ion must cancel the downward field due to the two positive ions. This gives (ignoring the factor of ${e}/{4 pi epsilon_0}$)

$$frac{1}{(y+sqrt{3})^2} = 2 cdot frac{1}{1+y^2} left( frac{y}{sqrt{1+y^2}} right)$$

where the $y/ sqrt{1 + y^2}$ factor in the first equality arises from taking the vertical component of the titled field lines due to the positive ions."
This equation has to be solved numerically and there is only one solution, which turns out to be $yapprox 0.1463$.

To find the other point (beyond the negative ion) where the force vanishes, all we have to do is to exchange the $sqrt{3}$ by $-sqrt{3}$ in the above equation and it is claimed that the solution now is $y_1approx 6.2045$ but it seems to me there are two other solutions to this equation, namely $y_2 = sqrt{3}$ and $y_3 approx 0.2486$.

I can accept that there are multiple solutions to the equation but how can you figure out that $y_1$ is the one that makes sense from the point of view of physics?