x^(3)=7x-2

solve for x

asked by guest
on Oct 23, 2024 at 3:29 am



You asked:

Solve the equation \[{x}^{3} = 7 x - 2\] for the unknown \(x\).

MathBot Answer:

The 3 solutions to the equation are: \[\begin{aligned}x &= - \frac{2 \sqrt{21} \cos{\left(\frac{\arctan{\left(\frac{2 \sqrt{237}}{9} \right)}}{3} \right)}}{3} \approx -2.7784571\\x &= - \frac{\sqrt{7} \sin{\left(\frac{\arctan{\left(\frac{2 \sqrt{237}}{9} \right)}}{3} \right)}}{2} - 7 \operatorname{re}{\left(\frac{1}{\left(- \frac{1}{2} - \frac{\sqrt{3} i}{2}\right) \sqrt[3]{27 + 6 \sqrt{237} i}}\right)} + \frac{\sqrt{21} \cos{\left(\frac{\arctan{\left(\frac{2 \sqrt{237}}{9} \right)}}{3} \right)}}{6} + i \left(- 7 \operatorname{im}{\left(\frac{1}{\left(- \frac{1}{2} - \frac{\sqrt{3} i}{2}\right) \sqrt[3]{27 + 6 \sqrt{237} i}}\right)} + \frac{\sqrt{21} \sin{\left(\frac{\arctan{\left(\frac{2 \sqrt{237}}{9} \right)}}{3} \right)}}{6} + \frac{\sqrt{7} \cos{\left(\frac{\arctan{\left(\frac{2 \sqrt{237}}{9} \right)}}{3} \right)}}{2}\right) \approx 0.28916855 + 2.0 \cdot 10^{-142} i\\x &= \frac{\sqrt{7} \sin{\left(\frac{\arctan{\left(\frac{2 \sqrt{237}}{9} \right)}}{3} \right)}}{2} + \frac{\sqrt{21} \cos{\left(\frac{\arctan{\left(\frac{2 \sqrt{237}}{9} \right)}}{3} \right)}}{6} - 7 \operatorname{re}{\left(\frac{1}{\left(- \frac{1}{2} + \frac{\sqrt{3} i}{2}\right) \sqrt[3]{27 + 6 \sqrt{237} i}}\right)} + i \left(- \frac{\sqrt{7} \cos{\left(\frac{\arctan{\left(\frac{2 \sqrt{237}}{9} \right)}}{3} \right)}}{2} + \frac{\sqrt{21} \sin{\left(\frac{\arctan{\left(\frac{2 \sqrt{237}}{9} \right)}}{3} \right)}}{6} - 7 \operatorname{im}{\left(\frac{1}{\left(- \frac{1}{2} + \frac{\sqrt{3} i}{2}\right) \sqrt[3]{27 + 6 \sqrt{237} i}}\right)}\right) \approx 2.4892886 + 8.0 \cdot 10^{-143} i\end{aligned}\]


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