Compute the sum of the roots of the equation \( v\sqrt{v} - 4v + 5\sqrt{v} - 6 = 0 \) given that all of the roots are positive.

["Understanding and Compute the Sum of the Roots for the Equation ( v\sqrt{v} - 4v + 5\sqrt{v} - 6 = 0 )", "When solving equations involving radicals, transformations and substitutions often simplify the process and reveal deeper algebraic structure. This article explores how to compute the sum of the roots for the non-linear equation:", "[\nv\sqrt{v} - 4v + 5\sqrt{v} - 6 = 0\n]", "Given Condition: All roots are positive real numbers.", "---", "### Step 1: Substitution to Simplify the Equation", "The presence of both ( v ) and ( \sqrt{v} ) suggests a substitution based on ( \sqrt{v} ). Let:", "[\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 equation transforms it into a cubic polynomial in ( x ):", "[\nx^3 - 4x^2 + 5x - 6 = 0\n]", "---", "### Step 2: Finding the Sum of ( \sqrt{v} ) Roots", "Since ( x = \sqrt{v} ) and all roots ( v > 0 ) are positive, then ( x > 0 ), so only positive real roots ( x ) are valid. The sum of the roots of the cubic equation:", "[\nx^3 - 4x^2 + 5x - 6 = 0\n]", "can be found using Vieta’s formulas, which relate coefficients to sums and products of roots.", "For a cubic equation ( ax^3 + bx^2 + cx + d = 0 ), the sum of the roots is:", "[\n-\frac{b}{a}\n]", "Here, ( a = 1 ), ( b = -4 ), so:", "[\n\ ext{Sum of } x\ ext{'s} = -\frac{-4}{1} = 4\n]", "That is, the sum of the square roots of the original variable ( v ) is:", "[\nx_1 + x_2 + x_3 = 4\n]", "---", "### Step 3: Relating Roots in ( v ) to Roots in ( x )", "The original variable is ( v = x^2 ), so the roots in ( v ) are ( v_i = x_i^2 ). We seek the sum:", "[\n\ ext{Sum of roots in } v = x_1^2 + x_2^2 + x_3^2\n]", "We use the identity:", "[\nx_1^2 + x_2^2 + x_3^2 = (x_1 + x_2 + x_3)^2 - 2(x_1x_2 + x_2x_3 + x_3x_1)\n]", "From Vieta’s formulas:", "- Sum of roots: ( x_1 + x_2 + x_3 = 4 )\n- Sum of product of roots two at a time: ( x_1x_2 + x_2x_3 + x_3x_1 = \frac{c}{a} = 5 )", "Therefore:", "[\nx_1^2 + x_2^2 + x_3^2 = 4^2 - 2 \cdot 5 = 16 - 10 = 6\n]", "---", "### Final Answer", "The sum of the positive roots ( v ) of the equation is:", "[\n\boxed{6}\n]", "---", "### Summary", "- Transformed the original radical equation via substitution ( x = \sqrt{v} ), yielding a cubic in ( x ).\n- Used Vieta’s formulas to find the sum of ( x )-roots: ( 4 ).\n- Computed the sum of squares of roots in ( x ) to obtain the sum in ( v ), resulting in ( 6 ).\n- Verified that all roots are positive, satisfying the given condition.", "This approach efficiently solves the equation and computes the desired sum using algebraic substitutions and foundational polynomial identities."]