Question: A science policy analyst studying energy output models a system where total power $ P(x) = x\sqrt{x} - 5x + 6\sqrt{x} $ is produced based on input $ x \geq 0 $. Find the sum of all real values of $ x $ for which $ P(x) = 0 $, given all roots are non-negative real numbers.

["Question: A science policy analyst studying energy output models a system where total power $ P(x) = x\sqrt{x} - 5x + 6\sqrt{x} $ is produced based on input $ x \geq 0 $. Find the sum of all real values of $ x $ for which $ P(x) = 0 $, given all roots are non-negative real numbers.", "---", "Understanding the Equation\nWe are given the power function:\n$$\nP(x) = x\sqrt{x} - 5x + 6\sqrt{x}\n$$\nand we are to find the sum of all real, non-negative solutions to $ P(x) = 0 $. Since $ x \geq 0 $, the square root $ \sqrt{x} $ is defined and real, making this a valid domain.", "To simplify, let $ u = \sqrt{x} $. Then $ x = u^2 $, and $ x\sqrt{x} = u^2 \cdot u = u^3 $. Substituting into the equation:\n$$\nP(x) = u^3 - 5u^2 + 6u\n$$\nSo the equation $ P(x) = 0 $ becomes:\n$$\nu^3 - 5u^2 + 6u = 0\n$$", "---", "Factor the Polynomial\nFactor out $ u $:\n$$\nu(u^2 - 5u + 6) = 0\n$$\nNow factor the quadratic:\n$$\nu(u - 2)(u - 3) = 0\n$$", "Thus, the solutions are $ u = 0 $, $ u = 2 $, and $ u = 3 $.", "Since $ u = \sqrt{x} \geq 0 $, all three solutions are valid in the transformed domain.", "---", "Back-Substitute to Find $ x $\nRecall $ x = u^2 $, so the corresponding values of $ x $ are:\n- $ u = 0 \Rightarrow x = 0^2 = 0 $\n- $ u = 2 \Rightarrow x = 2^2 = 4 $\n- $ u = 3 \Rightarrow x = 3^2 = 9 $", "All values $ x = 0, 4, 9 $ satisfy $ x \geq 0 $, so they are acceptable.", "---", "Sum of All Real Solutions\n$$\nx = 0 + 4 + 9 = 13\n$$", "---", "Final Answer:\nThe sum of all real values of $ x \geq 0 $ satisfying $ P(x) = 0 $ is $ \boxed{13} $."]