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Home » \( b^2 + 64 = 100 \) - Groen Casting

\( b^2 + 64 = 100 \) - Groen Casting

📅 March 1, 2026 👤 scraface

Mar 01, 2026

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📌 \Delta = -4\left(-\frac{407}{3}\right)^3 - 27\left(-\frac{158}{3}\right)^2 = -4\left(-\frac{67432643}{27}\right) - 27\cdot\frac{24964}{9}

📌 Besser: positives Vorzeichen:

📌 \( -4p^3 = -4 \cdot \left( -\frac{407^3}{27} \right) = \frac{4}{27} \cdot 67432643 \), huge positive

📌 \( 27q^2 = 27 \cdot \frac{24964}{9} = 3 \cdot 24964 = 74892 \)

📌 But \( p \) is large negative denominator: \( p = -\frac{407}{3} \Rightarrow p^3 = -\frac{67432643}{27} \), so \( -4p^3 = -4 \cdot (-\frac{67432643}{27}) = \frac{269730372}{27} \approx 9996788 \)

📌 \( q^2 = \left(\frac{158}{3}\right)^2 = \frac{24964}{9} \), so \( 27q^2 = 27 \cdot \frac{24964}{9} = 3 \cdot 24964 = 74892 \)

📌 Thus \( \Delta = 9996788 - 74892 = 9920896 > 0 \), so **three distinct real roots**!

📌 So all three roots \( w_1, w_2, w_3 \) are real. But earlier evaluations showed only one sign change. Contradiction?

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