Answer: \(k\gt \dfrac18\).

We start with the main method. Use the discriminant to decide how many real roots the quadratic equation has.

For the quadratic \(kx^2+(2k-1)x+k+1=0\) to have no real roots, its discriminant must be negative.

Here \(a=k\), \(b=2k-1\), and \(c=k+1\). So

\(\Delta=(2k-1)^2-4k(k+1).\)

Simplifying,

\(\Delta=4k^2-4k+1-4k^2-4k=1-8k.\)

Now require

\(1-8k\lt 0.\)

Thus \(8k\gt 1\), so

\(k\gt \frac18.\)

This completes the solution and gives the required result.