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📅 March 1, 2026 👤 scraface

Mar 01, 2026

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Solution: Let $ \sqrt{x} = y $, so $ x = y^2 $. Substituting into the...
y^2 \cdot y - 5y^2 + 6y = 0 \Rightarrow y^3 - 5y^2 + 6y = 0
Factor out $ y $:
y(y^2 - 5y + 6) = 0 \Rightarrow y(y - 2)(y - 3) = 0
So the solutions for $ y $ are $ y = 0, 2, 3 $. Since $ x = y^2 $, the...
$ y = 0 \Rightarrow x = 0 $
$ y = 2 \Rightarrow x = 4 $
$ y = 3 \Rightarrow x = 9 $
All roots are non-negative. The sum of the roots is $ 0 + 4 + 9 = 13 $...
Question: An industrial designer is modeling the cross-section of a su...

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