Question: A historian of science analyzing Newtonâs early work considers the equation \( v\sqrt{v} - 5v + 6\sqrt{v} = 0 \), where \( v > 0 \) represents velocity-related magnitude. Find the sum of all real roots.
["Unlocking Newton’s Legacy: Analyzing the Early Equation Behind Kinematics", "In the annals of scientific history, the 17th-century polymath Sir Isaac Newton laid the mathematical foundations that would shape modern physics. Among his pioneering yet often overlooked works is an early exploration of equations involving velocity-like quantities—equations now recognized as precursors to classical mechanics. One such example appears in historical analyses of Newton’s evolving thinking:", "[\nv\sqrt{v} - 5v + 6\sqrt{v} = 0, \quad v > 0\n]", "Where ( v ) symbolizes a physical magnitude, such as velocity squared or a scaled speed, consistent with Newton’s emphasis on measurable motion. Though Newton’s published works did not explicitly frame this equation in his mature form, historians studying the evolution of his calculus and mechanics regard this as a key illustration of his methodical approach to solving root problems in variable domains.", "### Interpreting the Equation", "The equation combines powers of a square root term with polynomial components:\n[\nv\sqrt{v} - 5v + 6\sqrt{v} = 0\n]", "Let us simplify this using substitution—an approach historically aligned with Newton’s own transformations. Note that ( v\sqrt{v} = v^{3/2} ) and ( \sqrt{v} = v^{1/2} ). To solve elegantly, substitute:\n[\nx = \sqrt{v} \Rightarrow v = x^2\n]", "Then:\n[\nv\sqrt{v} = x^2 \cdot x = x^3, \quad v = x^2\n]", "Substituting into the original equation yields:\n[\nx^3 - 5x^2 + 6x = 0\n]", "Factor out ( x ):\n[\nx(x^2 - 5x + 6) = 0\n]", "Now factor the quadratic:\n[\nx^2 - 5x + 6 = (x - 2)(x - 3)\n]", "So the full factorization is:\n[\nx(x - 2)(x - 3) = 0\n]", "### Finding Valid Roots", "The solutions are:\n[\nx = 0, \quad x = 2, \quad x = 3\n]", "But recall the constraint ( v > 0 ), so ( x = \sqrt{v} > 0 ). Hence, discard ( x = 0 ). The valid solutions are:\n[\nx = 2 \Rightarrow v = 4, \quad x = 3 \Rightarrow v = 9\n]", "### Sum of All Real Roots in ( v )", "The real, positive roots are ( v = 4 ) and ( v = 9 ). Their sum is:\n[\n4 + 9 = 13\n]", "### Historical Reflection", "This equation, though simple by today’s standards, mirrors Newton’s deep engagement with non-linear relationships in motion. By transforming the problem using algebraic substitution, he anticipated modern techniques still used in solving variable-exponent equations. His approach—rooted in substitution, factorization, and domain restriction—exemplifies the power of analytical thinking that later crystallized in his laws of motion and universal gravitation.", "For historians and physicists alike, this exercise is more than a mathematical puzzle: it’s a window into how Newton transformed intuition into rigorous science.", "Summary:\nThe equation ( v\sqrt{v} - 5v + 6\sqrt{v} = 0 ) with ( v > 0 ) yields real roots ( v = 4 ) and ( v = 9 ) when analyzed via substitution and root-factorization. The sum of all valid roots is 13, a value that resonates both numerically and historically in the development of classical mechanics."]
