Question: A quantum diagnostic device uses a function $ f(u) = u\sqrt{u} - 5u + 6\sqrt{u} $. Find the sum of all real roots of the equation $ f(u) = 0 $, given that all roots are non-negative.
["Title:\nSolve $ f(u) = u\sqrt{u} - 5u + 6\sqrt{u} = 0 $: Find the Sum of All Real Non-Negative Roots", "Meta Description:\nExplore the quantum diagnostic function $ f(u) = u\sqrt{u} - 5u + 6\sqrt{u} $. Learn how to find the sum of all real non-negative roots using substitution and algebraic techniques. Ideal for students and professionals in quantum biology and diagnostic modeling.", "---", "### Introduction", "In advanced quantum diagnostic models, non-linear functional equations arise when analyzing subatomic particle behavior under measurement. One such equation, relevant to signal resolution thresholds, is:\n[\nf(u) = u\sqrt{u} - 5u + 6\sqrt{u} = 0,\n]\nwhere $ u \geq 0 $ due to the presence of square roots. This article guides you step by step through solving for $ u $, transforming the equation into a polynomial form, and computing the sum of all real non-negative roots.", "---", "### Step 1: Substitution to Simplify the Equation", "The function contains terms involving $ \sqrt{u} $, suggesting a substitution to eliminate square roots. Let:\n[\n\sqrt{u} = x \quad \ ext{where } x \geq 0.\n]\nThen $ u = x^2 $, and $ u\sqrt{u} = x^2 \cdot x = x^3 $. Substitute into $ f(u) $:\n[\nf(u) = x^3 - 5x^2 + 6x = 0.\n]", "This is now a cubic equation in $ x $, with $ x \geq 0 $.", "---", "### Step 2: Factor the Polynomial", "Factor the equation:\n[\nx^3 - 5x^2 + 6x = 0\n]\nFactor out $ x $:\n[\nx(x^2 - 5x + 6) = 0.\n]\nNow factor the quadratic:\n[\nx^2 - 5x + 6 = (x - 2)(x - 3).\n]\nSo the full factorization is:\n[\nx(x - 2)(x - 3) = 0.\n]", "---", "### Step 3: Solve for $ x $", "Set each factor equal to zero:\n[\nx = 0, \quad x = 2, \quad x = 3.\n]\nAll solutions are non-negative, as required.", "---", "### Step 4: Convert Back to $ u $", "Recall $ u = x^2 $. Then the corresponding values of $ u $ are:\n- $ x = 0 \Rightarrow u = 0^2 = 0 $,\n- $ x = 2 \Rightarrow u = 2^2 = 4 $,\n- $ x = 3 \Rightarrow u = 3^2 = 9 $.", "Thus, the real non-negative roots are $ u = 0, 4, 9 $.", "---", "### Step 5: Compute the Sum of Roots", "Add the roots:\n[\n0 + 4 + 9 = 13.\n]", "---", "### Conclusion", "The equation $ u\sqrt{u} - 5u + 6\sqrt{u} = 0 $, representing a quantum diagnostic tuning condition, has three real non-negative roots: $ 0, 4, 9 $. The sum of all real roots is therefore:\n[\n\boxed{13}\n]\nThis algebraic insight supports precise calibration in quantum-based biological measurement systems.", "---", "### Key Takeaways\n- Used substitution $ \sqrt{u} = x $ to convert a radical equation into a cubic.\n- Factored completely to find all real roots in the domain $ u \geq 0 $.\n- Computed the sum of roots, verifying results via algebraic validation.\n- Highlighted application in quantum diagnostics modeling.", "For deeper insights into pattern recognition in quantum functional equations, explore variable substitution and transformation techniques.", "---", "Keywords: quantum diagnostic function, $ f(u) = u\sqrt{u} - 5u + 6\sqrt{u} $, sum of real roots, radical equation solution, substitution method, polynomial factoring, $ \sqrt{u} $ substitution, sum of roots in quantum modeling."]
