Answer: \(\displaystyle k=\frac{1+\sqrt{13}}{6}\) or \(\displaystyle k=\frac{1-\sqrt{13}}{6}\).

At a tangent point, the line and the curve meet at one repeated point. After equating the two expressions for \(y\), use the discriminant condition \(b^2-4ac=0\).

At a point of intersection, the line and the curve have the same \(y\)-value:

\(kx^2+x+2k=1-k-x.\)

Rearranging gives

\(kx^2+2x+3k-1=0.\)

For the line to be a tangent to the curve, this quadratic in \(x\) must have exactly one repeated root. Therefore its discriminant is zero:

\(2^2-4(k)(3k-1)=0.\)

So

\(4=4k(3k-1).\)

Dividing by 4,

\(1=3k^2-k.\)

Hence

\(3k^2-k-1=0.\)

Using the quadratic formula,

\(k=\frac{1\pm\sqrt{(-1)^2-4(3)(-1)}}{2(3)} =\frac{1\pm\sqrt{13}}{6}.\)