Home / Solution: Let $ \sqrt{x} = y $, so $ x = y^2 $. Substituting into the equation gives:

Solution: Let $ \sqrt{x} = y $, so $ x = y^2 $. Substituting into the equation gives:

February 22, 2026

$Solution: Let $ \sqrt{x} = y $, so $ x = y^2 $. Substituting into the equation gives:$

Substituting $s = 6$, we have:
R = \frac{6}{\sqrt{3}} = 2\sqrt{3}
Therefore, the radius of the circumscribed circle is $\boxed{2\sqrt{3}}$.Question: A computational biologist is analyzing a dataset of genetic sequences and encounters the equation $ x\sqrt{x} - 5x + 6\sqrt{x} = 0 $. Compute the sum of all real, non-negative roots of this equation.
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