Question: Compute the sum of the roots of the equation \( v\sqrt{v} - 5v + 6\sqrt{v} = 0 \), given that all roots are non-negative real numbers.

["Title: How to Compute the Sum of the Roots of the Equation ( v\sqrt{v} - 5v + 6\sqrt{v} = 0 ) – A Complete Guide", "---", "Meta Description:\nLearn how to compute the sum of the non-negative real roots of the equation ( v\sqrt{v} - 5v + 6\sqrt{v} = 0 ). Discover substitution methods, algebraic transformations, and Vieta’s insights for solving equations involving radicals.", "---", "## Understanding the Equation: ( v\sqrt{v} - 5v + 6\sqrt{v} = 0 )", "The equation\n[\nv\sqrt{v} - 5v + 6\sqrt{v} = 0\n]\ncontains mixed powers of ( v ) and its square root ( \sqrt{v} ), making it nonlinear and non-polynomial. However, it can be simplified effectively using substitution—especially since all roots are specified to be non-negative real numbers.", "---", "### Step 1: Substitution to Simplify the Radical", "Since ( \sqrt{v} ) appears repeatedly, we use substitution to eliminate the radical:", "Let\n[\nx = \sqrt{v} \quad \Rightarrow \quad v = x^2 \quad \ ext{and} \quad v\sqrt{v} = x^2 \cdot x = x^3\n]", "Substituting into the original equation:\n[\nx^3 - 5x^2 + 6x = 0\n]", "This is now a cubic polynomial equation in ( x ), with only real and non-negative roots (since ( v \geq 0 \Rightarrow x \geq 0 )).", "---", "### Step 2: Factor the Polynomial", "Factor out the common term ( 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]", "---", "### Step 3: Solve for ( x )", "Setting each factor to zero:\n[\nx = 0, \quad x = 2, \quad x = 3\n]", "All roots are non-negative, as required.", "---", "### Step 4: Convert Back to ( v )", "Recall ( v = x^2 ), so convert each root:", "- ( x = 0 \Rightarrow v = 0^2 = 0 )\n- ( x = 2 \Rightarrow v = 2^2 = 4 )\n- ( x = 3 \Rightarrow v = 3^2 = 9 )", "Thus, the non-negative real roots are ( v = 0, 4, 9 )", "---", "### Step 5: Compute the Sum of the Roots", "[\n0 + 4 + 9 = 13\n]", "---", "## Why This Method Works: The Power of Substitution", "By letting ( x = \sqrt{v} ), we transformed a radical equation into a manageable cubic, leveraging algebra to reveal structure. This substitution is a fundamental technique for solving equations with mixed powers involving square roots.", "---", "## Final Insight: Vieta’s Formula (Optional Advanced Check)", "Although Vieta’s formula applies directly to polynomial roots, here we used factorization to confirm the sum. For a cubic ( x^3 - 5x^2 + 6x = 0 ), the full sum of roots (including multiplicity and all sign combinations) is 5 (from coefficient of ( x^2 )). However, only non-negative roots contribute here — and indeed, ( 0 + 2 + 3 = 5 ), verifying consistency.", "---", "### Conclusion", "The sum of the non-negative real roots of ( v\sqrt{v} - 5v + 6\sqrt{v} = 0 ) is ( 13 ), found efficiently by substituting ( x = \sqrt{v} ), solving for ( x ), converting back, and summing.", "This method demonstrates how strategic substitution simplifies challenging radical equations—key for algebra students and math enthusiasts alike.", "---", "Keywords: compute sum of roots, equation ( v\sqrt{v} - 5v + 6\sqrt{v} = 0 ), radical equation, substitution method, sum of roots, algebra tutorial, solving equations involving radicals, ( x = \sqrt{v} )", "Frameworks optimized:\n- SEO title & meta description with primary keywords\n- Clear section hierarchy\n- Actionable steps with logical progression\n- Bonus insights with Vieta’s connection\n- Fully accessible to learners, search engines, and educators", "---", "Ready to solve more equations? Try this pattern: substitution, simplify, solve — success waiting."]