Since $ t = \sqrt{u} \geq 0 $, and all roots are real and non-negative, each root $ t $ corresponds to a non-negative $ u = t^2 $. Therefore, the sum of the values of $ u $ is not directly the sum of $ t^2 $, but the question asks for the **sum of the roots of the original equation in $ u $**, which are $ t^2 $, so we compute:

["Understanding the Sum of Roots in $ u $: A Closer Look", "In equations involving square roots, transforming variables is often key to revealing hidden structure. Consider the equation in terms of $ t $, where $ t = \sqrt{u} \geq 0 $: since $ u = t^2 $, and all roots $ t $ are real and non-negative, each valid solution $ t $ maps directly to a corresponding non-negative root in $ u $ as $ u = t^2 $.", "A common misconception is to sum the $ t $-values directly when asked for the “sum of the roots of the original equation in $ u $.” However, this overlooks the transformation. Since each root $ u $ in the original equation satisfies $ u = t^2 $, the roots in $ u $ are not $ t $, but rather the squares of the $ t $-roots.", "Let $ t_1, t_2, \dots, t_n $ be the non-negative real roots of the equation in $ t $, where $ t_i \geq 0 $. Then the corresponding roots in $ u $ are $ u_i = t_i^2 $. The sum of the roots in $ u $ is therefore:", "[\n\sum_{i=1}^{n} u_i = \sum_{i=1}^{n} t_i^2\n]", "Importantly, this sum is not equal to $ \sum t_i $, nor can it be simplified without knowledge of each individual $ t_i $. It only becomes computable if the full set of $ t $-roots is known explicitly.", "Suppose the equation after substitution becomes a polynomial in $ t $, say of degree $ n $, with all real, non-negative roots. Then, using symmetric sums and identities such as:", "[\n\sum t_i^2 = \left( \sum t_i \right)^2 - 2 \sum_{i < j} t_i t_j\n]", "we can compute $ \sum t_i^2 $ using known coefficients — provided the polynomial is fully specified.", "For example, consider a simple case: $ \sqrt{u} = t \Rightarrow u = t^2 $. If the original equation in $ t $ is $ t^2 - 5t + 6 = 0 $, the roots are $ t = 2, 3 $. Then the corresponding $ u $-roots are $ u = 4 $ and $ u = 9 $. The sum of $ u $-roots is $ 4 + 9 = 13 $. Alternatively, using identity:", "[\n\sum t_i^2 = 2^2 + 3^2 = 4 + 9 = 13\n]", "Or via coefficients: $ (\sum t_i)^2 - 2 \sum t_i t_j = 5^2 - 2(6) = 25 - 12 = 13 $", "This illustrates a powerful method: transform the equation in $ t $, compute $ \sum t_i^2 $ using symmetric sums, and convert to $ \sum u_i $ accordingly.", "In summary, when asked for the sum of the roots of the original equation in $ u $, convert each $ t $-root $ t_i $ to $ u_i = t_i^2 $, then sum:", "[\n\sum u_i = \sum t_i^2 = \left( \sum t_i \right)^2 - 2 \sum t_i t_j\n]", "This approach ensures accurate results grounded in algebraic principles, avoiding pitfalls of direct linear summation.", "Key takeaway: Always map transformations explicitly — the sum of $ u $-roots is not the square of the sum of $ t $-roots, but the sum of squares of $ t $-roots, derived from symmetric polynomial identities.", "By mastering this transformation logic, you unlock deeper insight into root behavior across variable substitutions — essential for solving and analyzing radical equations with precision."]