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