2x^3 + 13x^2 - 40x + 32 = 0

asked by guest
on Oct 24, 2024 at 7:19 am



You asked:

Solve the equation \(2 \cdot {x}^{3} + 13 \cdot {x}^{2} - 40 x + 32 = 0\) for the variable \(x\).

MathBot Answer:

The 3 solutions to the equation are: \[\begin{aligned}x &= - \frac{13}{6} - \frac{409}{12 \sqrt[3]{\frac{8605}{8} + 3 \sqrt{9771}}} - \frac{\sqrt[3]{\frac{8605}{8} + 3 \sqrt{9771}}}{3} \approx -8.9379366\\x &= - \frac{13}{6} + \frac{409}{24 \sqrt[3]{3 \sqrt{9771} + \frac{8605}{8}}} + \frac{\sqrt[3]{3 \sqrt{9771} + \frac{8605}{8}}}{6} + i \left(- \frac{409 \sqrt{3}}{24 \sqrt[3]{3 \sqrt{9771} + \frac{8605}{8}}} + \frac{\sqrt{3} \sqrt[3]{3 \sqrt{9771} + \frac{8605}{8}}}{6}\right) \approx 1.2189683 + 0.55157835 i\\x &= - \frac{13}{6} + \frac{409}{24 \sqrt[3]{3 \sqrt{9771} + \frac{8605}{8}}} + \frac{\sqrt[3]{3 \sqrt{9771} + \frac{8605}{8}}}{6} + i \left(- \frac{\sqrt{3} \sqrt[3]{3 \sqrt{9771} + \frac{8605}{8}}}{6} + \frac{409 \sqrt{3}}{24 \sqrt[3]{3 \sqrt{9771} + \frac{8605}{8}}}\right) \approx 1.2189683 -0.55157835 i\end{aligned}\]

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

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