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Solve for z.

z {}^(4) - 65z {}^(2) + 64 = 0

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User DarkCrow
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1 Answer

4 votes

\bf z^4-65z^2+64=0\implies (z^2)^2-65(z^2)+64=0 \\\\\\ (z^2-64)(z^2-1)=0\implies \begin{cases} z^2-64=0\\ z^2=64\\ z=\pm√(64)\\ z=\pm 8\\ ------\\ z^2-1=0\\ z^2=1\\ z=\pm√(1)\\ z=\pm 1 \end{cases}
answered
User Peter Jamieson
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8.1k points

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