M^4+M^2+16=0

asked by guest
on Sep 20, 2024 at 5:43 am



You asked:

Solve the equation \({M}^{4} + {M}^{2} + 16 = 0\) for the variable \(M\).

MathBot Answer:

The 4 solutions to the equation are: \[\begin{aligned}M &= - 2 \sin{\left(\frac{\arctan{\left(3 \sqrt{7} \right)}}{2} \right)} - 2 i \cos{\left(\frac{\arctan{\left(3 \sqrt{7} \right)}}{2} \right)} \approx -1.3228757 -1.5 i\\M &= - 2 \sin{\left(\frac{\arctan{\left(3 \sqrt{7} \right)}}{2} \right)} + 2 i \cos{\left(\frac{\arctan{\left(3 \sqrt{7} \right)}}{2} \right)} \approx -1.3228757 + 1.5 i\\M &= 2 \sin{\left(\frac{\arctan{\left(3 \sqrt{7} \right)}}{2} \right)} - 2 i \cos{\left(\frac{\arctan{\left(3 \sqrt{7} \right)}}{2} \right)} \approx 1.3228757 -1.5 i\\M &= 2 \sin{\left(\frac{\arctan{\left(3 \sqrt{7} \right)}}{2} \right)} + 2 i \cos{\left(\frac{\arctan{\left(3 \sqrt{7} \right)}}{2} \right)} \approx 1.3228757 + 1.5 i\end{aligned}\]

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

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