1. Problem 16: What is $i^2$? Recall that $i$ is the imaginary unit defined by $i = \sqrt{-1}$. By definition, $i^2 = -1$. Answer: C) -1 2. Problem 17: Simplify $\sqrt{-25}$. We can write $\sqrt{-25} = \sqrt{25 \times -1} = \sqrt{25} \times \sqrt{-1} = 5i$. Answer: D) 5i 3. Problem 18: Square the imaginary number $i\sqrt{5}$. Calculate: $$ (i\sqrt{5})^2 = i^2 \times (\sqrt{5})^2 = (-1) \times 5 = -5 $$ Answer: C) -5 4. Problem 19: Explain the error in Trent's solution. Trent's work: $$ 5x^2 + 1000 = -125 $$ $$ 5x^2 = -1125 $$ $$ x^2 = -225 $$ $$ x = \pm \sqrt{-225} $$ $$ x = \pm 15 $$ Error explanation: The mistake is in the last step. The square root of a negative number is not a real number. Specifically, $$ \sqrt{-225} = \sqrt{225 \times -1} = 15i $$ So the correct solution is: $$ x = \pm 15i $$ 5. Problem 20: Solve $x^2 = -4$. Rewrite: $$ x^2 = -4 $$ Take square root on both sides: $$ x = \pm \sqrt{-4} = \pm \sqrt{4 \times -1} = \pm 2i $$ Answer: A) \pm 2i