x^9 = 0.4975

asked by guest
on Oct 26, 2024 at 8:40 am



You asked:

Solve the equation \({x}^{9} = 0.4975\) for the variable \(x\).

MathBot Answer:

The solutions are given by the formula: \[x = \frac{\sqrt[9]{497500000} \cos{\left(\frac{2 \pi k}{9} \right)}}{10} + \frac{\sqrt[9]{497500000} \sin{\left(\frac{2 \pi k}{9} \right)}}{10} i\] where \(k\) is an integer in \(0 \le k < 9\).

There is \(1\) real solution.

There are \(8\) solutions with nonzero imaginary part.


Real solutions: \[\begin{aligned}x &= \frac{\sqrt[9]{497500000}}{10} \\&\approx 0.92535919\end{aligned}\]


Solutions with nonzero imaginary part: \[\begin{aligned}x &= \frac{\sqrt[9]{497500000} \cos{\left(\frac{2 \pi}{9} \right)}}{10} + \frac{\sqrt[9]{497500000} i \sin{\left(\frac{2 \pi}{9} \right)}}{10} \\&\approx 0.70886627 + 0.59480942 i\\x &= \frac{\sqrt[9]{497500000} \cos{\left(\frac{4 \pi}{9} \right)}}{10} + \frac{\sqrt[9]{497500000} i \sin{\left(\frac{4 \pi}{9} \right)}}{10} \\&\approx 0.16068694 + 0.91130091 i\\x &= - \frac{\sqrt[9]{497500000}}{20} + \frac{\sqrt{3} \cdot \sqrt[9]{497500000} i}{20} \\&\approx -0.4626796 + 0.80138457 i\\x &= - \frac{\sqrt[9]{497500000} \cos{\left(\frac{\pi}{9} \right)}}{10} + \frac{\sqrt[9]{497500000} i \sin{\left(\frac{\pi}{9} \right)}}{10} \\&\approx -0.8695532 + 0.31649148 i\\x &= - \frac{\sqrt[9]{497500000} \cos{\left(\frac{\pi}{9} \right)}}{10} - \frac{\sqrt[9]{497500000} i \sin{\left(\frac{\pi}{9} \right)}}{10} \\&\approx -0.8695532 -0.31649148 i\\x &= - \frac{\sqrt[9]{497500000}}{20} - \frac{\sqrt{3} \cdot \sqrt[9]{497500000} i}{20} \\&\approx -0.4626796 -0.80138457 i\\x &= \frac{\sqrt[9]{497500000} \cos{\left(\frac{4 \pi}{9} \right)}}{10} - \frac{\sqrt[9]{497500000} i \sin{\left(\frac{4 \pi}{9} \right)}}{10} \\&\approx 0.16068694 -0.91130091 i\\x &= \frac{\sqrt[9]{497500000} \cos{\left(\frac{2 \pi}{9} \right)}}{10} - \frac{\sqrt[9]{497500000} i \sin{\left(\frac{2 \pi}{9} \right)}}{10} \\&\approx 0.70886627 -0.59480942 i\end{aligned}\]


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