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📅 May 14, 2026 👤 scraface

May 14, 2026

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u\sqrt{u} - 5u + 6\sqrt{u} = 0,
given that all of the roots are real and nonnegative.
Solution: Let us make the substitution $ x = \sqrt{u} $, so that $ u =...
x^2 \cdot x - 5x^2 + 6x = 0 \quad \Rightarrow \quad x^3 - 5x^2 + 6x =...
Factor out $ x $:
x(x^2 - 5x + 6) = 0.
Now factor the quadratic:
x(x - 2)(x - 3) = 0.
So the solutions for $ x $ are $ x = 0, 2, 3 $. Since $ x = \sqrt{u} $...
u = 0^2 = 0, \quad u = 2^2 = 4, \quad u = 3^2 = 9.

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