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Home » $ f(ab) = f(a)f(b) $ for all real $ a, b $ - Groen Casting

$ f(ab) = f(a)f(b) $ for all real $ a, b $ - Groen Casting

📅 March 1, 2026 👤 scraface

Mar 01, 2026

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📌 Compute the sum of the roots of the equation \( u\sqrt{u} - 5u + 6\sqrt{u} = 0 \), given that all roots are non-negative.

📌 Let \( v = \sqrt{u} \), so \( u = v^2 \). Substitute:

📌 v^2 \cdot v - 5v^2 + 6v = 0 \quad \Rightarrow \quad v^3 - 5v^2 + 6v = 0.

📌 v(v^2 - 5v + 6) = 0.

📌 Factor quadratic:

📌 v(v - 2)(v - 3) = 0.

📌 So, \( v = 0, 2, 3 \). Since \( u = v^2 \), the corresponding roots are:

📌 All roots are non-negative, so valid.

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