5229.8=3.14D^2*3D

asked by guest
on Dec 30, 2024 at 2:36 pm



You asked:

Solve the equation \(5229.8 = 3.14 {D}^{2} \cdot 3 D\) for the variable \(D\).

MathBot Answer:

The solutions are given by the formula: \[D = \frac{471^{\frac{2}{3}} \cdot \sqrt[3]{261490} \cos{\left(\frac{2 \pi k}{3} \right)}}{471} + \frac{471^{\frac{2}{3}} \cdot \sqrt[3]{261490} \sin{\left(\frac{2 \pi k}{3} \right)}}{471} i\] where \(k\) is an integer in \(0 \le k < 3\).

There is \(1\) real solution.

There are \(2\) solutions with nonzero imaginary part.

Real solutions: \[\begin{aligned}D &= \frac{471^{\frac{2}{3}} \cdot \sqrt[3]{261490}}{471} \\&\approx 8.2188564\end{aligned}\]

Solutions with nonzero imaginary part: \[\begin{aligned}D &= - \frac{\sqrt[3]{261490} \cdot 471^{\frac{2}{3}}}{942} + \frac{157^{\frac{2}{3}} \cdot \sqrt[3]{261490} \cdot \sqrt[6]{3} i}{314} \\&\approx -4.1094282 + 7.1177384 i\\D &= - \frac{\sqrt[3]{261490} \cdot 471^{\frac{2}{3}}}{942} - \frac{157^{\frac{2}{3}} \cdot \sqrt[3]{261490} \cdot \sqrt[6]{3} i}{314} \\&\approx -4.1094282 -7.1177384 i\end{aligned}\]

\(i\) is the imaginary unit, defined as \(i^2 = -1\).

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