V^3 + 290.45V - 144100 = 0

asked by guest
on Jan 25, 2025 at 1:10 am



You asked:

Solve the equation \({V}^{3} + 290.45 V - 144100 = 0\) for the variable \(V\).

MathBot Answer:

The 3 solutions to the equation are: \[\begin{aligned}V &= \sqrt[3]{72050 + \frac{7 \sqrt{343315029088815}}{1800}} - \frac{5809}{60 \sqrt[3]{72050 + \frac{7 \sqrt{343315029088815}}{1800}}} \approx 50.581052\\V &= - \frac{\sqrt[3]{72050 + \frac{7 \sqrt{343315029088815}}{1800}}}{2} + \frac{5809}{120 \sqrt[3]{72050 + \frac{7 \sqrt{343315029088815}}{1800}}} + i \left(\frac{5809 \sqrt{3}}{120 \sqrt[3]{72050 + \frac{7 \sqrt{343315029088815}}{1800}}} + \frac{\sqrt{3} \sqrt[3]{72050 + \frac{7 \sqrt{343315029088815}}{1800}}}{2}\right) \approx -25.290526 + 47.003002 i\\V &= - \frac{\sqrt[3]{72050 + \frac{7 \sqrt{343315029088815}}{1800}}}{2} + \frac{5809}{120 \sqrt[3]{72050 + \frac{7 \sqrt{343315029088815}}{1800}}} + i \left(- \frac{\sqrt{3} \sqrt[3]{72050 + \frac{7 \sqrt{343315029088815}}{1800}}}{2} - \frac{5809 \sqrt{3}}{120 \sqrt[3]{72050 + \frac{7 \sqrt{343315029088815}}{1800}}}\right) \approx -25.290526 -47.003002 i\end{aligned}\]

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

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