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doubletree by hilton miami north i 95 - Imagemakers

📅 May 1, 2026 👤 scraface

May 01, 2026

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📌 x^3 - 4x^2 + 5x - 6 = 0

📌 Now, we need to find the sum of the roots of this cubic equation. In a cubic equation of the form \( ax^3 + bx^2 + cx + d = 0 \), the sum of the roots is given by:

📌 rac{b}{a} = -rac{-4}{1} = 4

📌 Since \( x = \sqrt{v} \), each positive root \( x_i \) corresponds to a positive root \( v_i = x_i^2 \). However, we are asked for the sum of the roots of the original equation in terms of \( v \), not \( x \). The sum of the roots of the original equation in \( v \) corresponds directly to the sum of \( x_i^2 \), but this is not simply the sum of the \( x_i \)'s. Instead, note that since we are only asked for the sum of roots (and given all are positive, and the transformation is valid), the number of valid \( x \)-roots translates to transformable \( v \)-roots, but the sum of the original \( v_i \) values corresponds to the sum of \( x_i^2 \), which is not directly \( 4^2 = 16 \).

📌 But wait: the sum of the roots in \( x \) is 4, but the sum of the roots in \( v \) is \( \sum x_i^2 \), not \( (\sum x_i)^2 \). Use the identity:

📌 \sum x_i^2 = \left( \sum x_i

📌 ight)^2 - 2\sum_{i<j} x_i x_j

📌 From the cubic \( x^3 - 4x^2 + 5x - 6 = 0 \), the symmetric sums are:

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