Compute the product of the roots of the equation \( v\sqrt{v} - 5v + 6\sqrt{v} = 0 \), given that all of the roots are real and positive.

["Compute the Product of the Roots of the Equation ( v\sqrt{v} - 5v + 6\sqrt{v} = 0 ): A Step-by-Step Guide", "When solving equations involving radicals, careful substitution can transform complex expressions into manageable polynomial forms—critical for accurately computing key properties like the product of roots. This article explores how to compute the product of the real, positive roots of the equation:", "[\nv\sqrt{v} - 5v + 6\sqrt{v} = 0\n]", "---", "### Step 1: Substitution to Simplify the Equation", "The presence of ( \sqrt{v} ) suggests a substitution will linearize the equation. Let:", "[\nx = \sqrt{v}\n]", "Then, since ( v = x^2 ) and ( \sqrt{v} = x ), substitute into the original equation:", "[\n(v\sqrt{v}) - 5v + 6\sqrt{v} = (x^2 \cdot x) - 5x^2 + 6x = x^3 - 5x^2 + 6x = 0\n]", "So the equation becomes:", "[\nx^3 - 5x^2 + 6x = 0\n]", "---", "### Step 2: Factor the Polynomial", "Factor out the common term ( x ):", "[\nx(x^2 - 5x + 6) = 0\n]", "Now factor the quadratic:", "[\nx(x - 2)(x - 3) = 0\n]", "Thus, the roots are:", "[\nx = 0, \quad x = 2, \quad x = 3\n]", "---", "### Step 3: Translate Back to ( v )", "Recall ( v = x^2 ), so the corresponding roots in ( v ) are:", "- ( x = 0 \Rightarrow v = 0^2 = 0 )\n- ( x = 2 \Rightarrow v = 2^2 = 4 )\n- ( x = 3 \Rightarrow v = 3^2 = 9 )", "However, the problem states that all roots are real and positive. Since ( v = 0 ) is not positive, we exclude it.", "Therefore, the valid roots are:", "[\nv = 4 \quad \ ext{and} \quad v = 9\n]", "---", "### Step 4: Compute the Product of Positive Roots", "Now compute the product of the valid roots:", "[\n4 \ imes 9 = 36\n]", "---", "### Why This Method Works", "Using substitution converts a mixed radical equation into a cubic polynomial, making root-finding straightforward. Because the problem specifies that all roots are real and positive, we selectively retain only those values of ( v > 0 ), ensuring accuracy in interpreting the physical or mathematical context—especially important in applied fields like engineering or physics.", "---", "### Final Answer", "[\n\boxed{36}\n]"]