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.