Solve the equation and each inequality. Use interval notation to write solution sets to the inequalities \( \begin{array}{lll}\text { (a) } k^{2}-11=0 & \text { (b) } k^{2}-11 \leq 0 & \text { (c) } k^{2}-11 \geq 0\end{array} \)

Let's solve each part of the problem step by step. ### (a) Solve the equation \( k^{2} - 11 = 0 \) To solve for \( k \), we can rearrange the equation: \[ k^{2} = 11 \] Taking the square root of both sides, we get: \[ k = \pm \sqrt{11} \] Thus, the solutions to the equation are: \[ k = \sqrt{11} \quad \text{and} \quad k = -\sqrt{11} \] ### (b) Solve the inequality \( k^{2} - 11 \leq 0 \) We can rearrange the inequality: \[ k^{2} \leq 11 \] Taking the square root of both sides gives us: \[ -\sqrt{11} \leq k \leq \sqrt{11} \] In interval notation, the solution set is: \[ [-\sqrt{11}, \sqrt{11}] \] ### (c) Solve the inequality \( k^{2} - 11 \geq 0 \) We can rearrange the inequality: \[ k^{2} \geq 11 \] Taking the square root of both sides gives us: \[ k \leq -\sqrt{11} \quad \text{or} \quad k \geq \sqrt{11} \] In interval notation, the solution set is: \[ (-\infty, -\sqrt{11}] \cup [\sqrt{11}, \infty) \] ### Summary of Solutions - (a) The solutions to the equation \( k^{2} - 11 = 0 \) are \( k = \sqrt{11} \) and \( k = -\sqrt{11} \). - (b) The solution to the inequality \( k^{2} - 11 \leq 0 \) is \( [-\sqrt{11}, \sqrt{11}] \). - (c) The solution to the inequality \( k^{2} - 11 \geq 0 \) is \( (-\infty, -\sqrt{11}] \cup [\sqrt{11}, \infty) \).