1. **State the problem:** Solve the equation $$\sqrt{2k+3} = k+2$$ and determine which of the statements I, II, and III are true. 2. **Solve the equation:** Square both sides to eliminate the square root: $$\left(\sqrt{2k+3}\right)^2 = (k+2)^2$$ $$2k + 3 = (k+2)^2$$ Expand the right side: $$2k + 3 = k^2 + 4k + 4$$ Bring all terms to one side: $$0 = k^2 + 4k + 4 - 2k - 3$$ $$0 = k^2 + 2k + 1$$ This factors as: $$(k + 1)^2 = 0$$ So, $$k = -1$$ 3. **Check for extraneous solutions:** Substitute $k = -1$ back into the original equation: $$\sqrt{2(-1) + 3} = \sqrt{1} = 1$$ $$k + 2 = -1 + 2 = 1$$ Both sides equal 1, so $k = -1$ is a valid solution. 4. **Evaluate the statements:** - I. $$k^k = k$$ Calculate: $$(-1)^{-1} = \frac{1}{-1} = -1$$ Right side: $$k = -1$$ So, $$(-1)^{-1} = -1$$ is true. - II. $$|k| = -k$$ Calculate: $$| -1 | = 1$$ $$-k = -(-1) = 1$$ So, $$|k| = -k$$ is true. - III. $$k^0 = -k$$ Calculate: $$k^0 = (-1)^0 = 1$$ $$-k = -(-1) = 1$$ So, $$k^0 = -k$$ is true. 5. **Conclusion:** All three statements I, II, and III are true. **Final answer:** (E) I, II and III