x^3 + x^2 - 4 = 0

asked by guest
on Nov 14, 2024 at 8:27 pm



You asked:

Solve the equation \({x}^{3} + {x}^{2} - 4 = 0\) for the variable \(x\).

MathBot Answer:

The 3 solutions to the equation are: \[\begin{aligned}x &= - \frac{1}{3} + \sqrt[3]{\frac{53}{27} + \frac{2 \sqrt{78}}{9}} + \frac{1}{9 \sqrt[3]{\frac{53}{27} + \frac{2 \sqrt{78}}{9}}} \approx 1.3145962\\x &= - \frac{\sqrt[3]{\frac{2 \sqrt{78}}{9} + \frac{53}{27}}}{2} - \frac{1}{3} - \frac{1}{18 \sqrt[3]{\frac{2 \sqrt{78}}{9} + \frac{53}{27}}} + i \left(- \frac{\sqrt{3}}{18 \sqrt[3]{\frac{2 \sqrt{78}}{9} + \frac{53}{27}}} + \frac{\sqrt{3} \sqrt[3]{\frac{2 \sqrt{78}}{9} + \frac{53}{27}}}{2}\right) \approx -1.1572981 + 1.3051515 i\\x &= - \frac{\sqrt[3]{\frac{2 \sqrt{78}}{9} + \frac{53}{27}}}{2} - \frac{1}{3} - \frac{1}{18 \sqrt[3]{\frac{2 \sqrt{78}}{9} + \frac{53}{27}}} + i \left(- \frac{\sqrt{3} \sqrt[3]{\frac{2 \sqrt{78}}{9} + \frac{53}{27}}}{2} + \frac{\sqrt{3}}{18 \sqrt[3]{\frac{2 \sqrt{78}}{9} + \frac{53}{27}}}\right) \approx -1.1572981 -1.3051515 i\end{aligned}\]

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

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