Let f (x) = ax 2+2x-1.

When a=0,

F(x)=2x- 1, let f(x)=0 and x= 1/2.

In other words, the original equation has a positive root.

When a>0, f (x) = ax 2+2x-1.

It is a quadratic function, and the parabolic opening of the image is upward.

∫f(0)=- 1 & lt； 0, axis of symmetry x =-1/a >; 0

∴f(x) image must intersect with the positive semi-axis of the x axis.

Meet the meaning of the question

When a< is 0,

The f(x) image must intersect with the positive semi-axis of the x axis.

Then δ = 4+4a ≥ 0 is required.

∴a≥- 1

∴- 1≤a<； 0

To sum up, there is at least one equation ax 2+2x-1= 0.

The necessary and sufficient condition for a positive real root is a≥- 1.