2x^3 -2 x^2 -x +2 = 0

asked by guest
on Nov 30, 2024 at 2:15 am



You asked:

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

MathBot Answer:

The 3 solutions to the equation are: \[\begin{aligned}x &= \frac{1}{3} - \frac{5}{6 \sqrt[3]{\frac{41}{4} + \frac{3 \sqrt{159}}{4}}} - \frac{\sqrt[3]{\frac{41}{4} + \frac{3 \sqrt{159}}{4}}}{3} \approx -0.87555035\\x &= \frac{5}{12 \sqrt[3]{\frac{3 \sqrt{159}}{4} + \frac{41}{4}}} + \frac{1}{3} + \frac{\sqrt[3]{\frac{3 \sqrt{159}}{4} + \frac{41}{4}}}{6} + i \left(- \frac{5 \sqrt{3}}{12 \sqrt[3]{\frac{3 \sqrt{159}}{4} + \frac{41}{4}}} + \frac{\sqrt{3} \sqrt[3]{\frac{3 \sqrt{159}}{4} + \frac{41}{4}}}{6}\right) \approx 0.93777518 + 0.51255877 i\\x &= \frac{5}{12 \sqrt[3]{\frac{3 \sqrt{159}}{4} + \frac{41}{4}}} + \frac{1}{3} + \frac{\sqrt[3]{\frac{3 \sqrt{159}}{4} + \frac{41}{4}}}{6} + i \left(- \frac{\sqrt{3} \sqrt[3]{\frac{3 \sqrt{159}}{4} + \frac{41}{4}}}{6} + \frac{5 \sqrt{3}}{12 \sqrt[3]{\frac{3 \sqrt{159}}{4} + \frac{41}{4}}}\right) \approx 0.93777518 -0.51255877 i\end{aligned}\]

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

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