According to the wikipedia article on octic equations, octic equations of the form $ax^8 \pm bx^4 \pm c = 0$ can be solved using the quadratic formula. How might one actually do this?
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3Substitute $x^4$ by $y$. The equation becomes $ay^2\pm by\pm c=0$ – callculus42 May 05 '16 at 19:29
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2Solve as a quadratic for $x^4$, then take the square root twice! – almagest May 05 '16 at 19:30
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Replacing $w=x^4$ it's familiar:
$$w = \frac{-b \pm \sqrt{b^2 \pm 4ac}}{2a}.$$
Then extract the eight roots by taking $v_1 = \sqrt{w}, v_2 = -\sqrt{w}$, and then $x=\pm \sqrt{v_1}$ and $x = \pm \sqrt{v_2}$, for each of the two values of $w$.
John
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So the values of X will be the positive and negative square roots of U1 and U2? – MrKagouris May 06 '16 at 08:57
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Just remember that there are two values of $w$ from the quadratic formula. You get four roots apiece for each value of $w$, which gives you eight roots (values for $x$). Some may be repeated roots. All of the complex roots will come in conjugate pairs. – John May 06 '16 at 17:04
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Oh, so I'd have to get u1 and u2 for one value of w but also u3 and u4 for the other value. Got it, thanks! – MrKagouris May 06 '16 at 18:48