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\[\boxed\frac\ln 35.\] - Appcentric

📅 February 27, 2026 👤 scraface

Feb 27, 2026

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given that all roots are real and non-negative.
Solution: Let $ \sqrt{u} = t $, so $ u = t^2 $. Substituting into the...
t^2 \cdot t - 6t^2 + 7t - 12 = 0 \Rightarrow t^3 - 6t^2 + 7t - 12 = 0
We are now to find the sum of the roots of the cubic equation
t^3 - 6t^2 + 7t - 12 = 0
By **Vieta’s Formula**, the sum of the roots is the coefficient of $ t...
\text{Sum of roots} = 6
Since $ t = \sqrt{u} \geq 0 $, and all roots are real and non-negative...
\sum u = \sum t^2
We compute $ \sum t^2 $ using the identity:

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