(i) Determine the set of values of \( k \) for which the equation \( x^{2}+2 x+k=3 k x-1 \) has no real roots.

Solve the equation \( x^{2}+2x+k=3kx-1 \). Solve the equation by following steps: - step0: Solve for \(x\): \(x^{2}+2x+k=3kx-1\) - step1: Move the expression to the left side: \(x^{2}+2x+k-\left(3kx-1\right)=0\) - step2: Remove the parentheses: \(x^{2}+2x+k-3kx+1=0\) - step3: Collect like terms: \(x^{2}+\left(2-3k\right)x+k+1=0\) - step4: Solve using the quadratic formula: \(x=\frac{-2+3k\pm \sqrt{\left(2-3k\right)^{2}-4\left(k+1\right)}}{2}\) - step5: Simplify the expression: \(x=\frac{-2+3k\pm \sqrt{-16k+9k^{2}}}{2}\) - step6: Separate into possible cases: \(\begin{align}&x=\frac{-2+3k+\sqrt{-16k+9k^{2}}}{2}\\&x=\frac{-2+3k-\sqrt{-16k+9k^{2}}}{2}\end{align}\) The solutions to the equation \(x^{2}+2x+k=3kx-1\) are given by the quadratic formula as: \[x=\frac{-2+3k\pm\sqrt{-16k+9k^{2}}}{2}\] To determine the set of values of \(k\) for which the equation has no real roots, we need to find the values of \(k\) for which the discriminant \(-16k+9k^{2}\) is negative. Let's solve the inequality \(-16k+9k^{2}<0\) to find the values of \(k\) for which the equation has no real roots. Solve the inequality by following steps: - step0: Solve the inequality by testing the values in the interval: \(-16k+9k^{2}<0\) - step1: Rewrite the expression: \(-16k+9k^{2}=0\) - step2: Factor the expression: \(-k\left(16-9k\right)=0\) - step3: Separate into possible cases: \(\begin{align}&-k=0\\&16-9k=0\end{align}\) - step4: Solve the equation: \(\begin{align}&k=0\\&k=\frac{16}{9}\end{align}\) - step5: Determine the test intervals: \(\begin{align}&k<0\\&0\frac{16}{9}\end{align}\) - step6: Choose a value: \(\begin{align}&k_{1}=-1\\&k_{2}=1\\&k_{3}=3\end{align}\) - step7: Test the chosen value: \(\begin{align}&k<0\textrm{ }\textrm{is not a solution}\\&0\frac{16}{9}\textrm{ }\textrm{is not a solution}\end{align}\) - step8: The final solution is \(0