Question 729832
 first re-arrange the equation base on {{{x}}}:
{{{y=(-1/m)*x}}} and {{{y=+sqrt(x)}}} or {{{y=-sqrt(x)}}} by writing the equilibrium for both sides base on "y" we would have:
{{{(-1/m)*x=sqrt(x)}}} then {{{(1/m^2)*x^2=x}}} then {{{(1/m^2)*x^2-x=0}}} then 
{{{x^2-(m^2)*x=0}}} then {{{x*(X-m^2)=0}}} that means:
{{{X=0}}} or {{{X=m^2}}} so, base on this answer it appears that regardless of what value {{{m}}} taken we always would have an intersection in the origin of {{{x=0}}} & {{{y=0}}}.
In addition, if we try to use a very small value for {{{m}}}, lets say for example {{{1/10^100000000}}} the second answer of {{{x}}} would be very close value to zero , In other words, as much as we take a smaller value for {{{m}}} we are more closer to a unique intersection value of zero, otherwise in bigger cases we can't neglect bigger values of {{{m}}} and the second intersection  would be at {{{x=m^2}}}.

{{{graph( 600, 400, -5, 5, -5, 5, -x, -2x, -10x,-sqrt(x),sqrt(x))}}}
In above graph the two curve line are {{{y=sqrt(x)}}} and {{{y=-sqrt(x)}}}
As much as the take a smaller value for {{{m}}} the straight line is going to get more vertical {(red m=1) (green m=0.2) (blue m= 0.1)}. so the second intersection point get closer more and more to zero.



In conclusion >>> 
1)for case b: we can't guarantee that for sure cause there will always be at least one intersection at {{{X=0}}}
2) for  case a if we try to minimize the value of {{{m}}} we are close to unique answer of one intersect at {{{X=0}}} but it's never gonna happen (try to draw both graphs to understand why)
3)for case c: it's what is always going to happen for the rest of non-small values of {{{m}}}.

hope that helps.