Haiphongum
2022-09-23
Answered

What is the process of solving an equation with the square and the 4th power of an unknown, for example ${x}^{4}-6{x}^{2}-27=0$

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Lilliana Mason

Answered 2022-09-24
Author has **11** answers

This is a quadratic in terms of ${x}^{2}$. I know solutions have already been posted, but this is the way I like to solve these problems:

${x}^{4}-6{x}^{2}-27=({x}^{2}-9)({x}^{2}+3)$

If we are solving: $({x}^{2}-9)({x}^{2}+3)=0$, this gives us two equations:

${x}^{2}-9=0\phantom{\rule{2em}{0ex}}\text{or}\phantom{\rule{2em}{0ex}}{x}^{2}+3=0$

The one on the left yields:

${x}^{2}=9\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}x=\pm 3$

The one on the right yields:(where i is the imaginary unit)

${x}^{2}=-3\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}x=\pm \sqrt{3}i$

These are the four roots promised us by the Fundamental Theorem of Algebra. And, as expected, the complex roots came in conjugate pairs.

${x}^{4}-6{x}^{2}-27=({x}^{2}-9)({x}^{2}+3)$

If we are solving: $({x}^{2}-9)({x}^{2}+3)=0$, this gives us two equations:

${x}^{2}-9=0\phantom{\rule{2em}{0ex}}\text{or}\phantom{\rule{2em}{0ex}}{x}^{2}+3=0$

The one on the left yields:

${x}^{2}=9\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}x=\pm 3$

The one on the right yields:(where i is the imaginary unit)

${x}^{2}=-3\phantom{\rule{thickmathspace}{0ex}}\u27f9\phantom{\rule{thickmathspace}{0ex}}x=\pm \sqrt{3}i$

These are the four roots promised us by the Fundamental Theorem of Algebra. And, as expected, the complex roots came in conjugate pairs.

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