There is a formula for cubic equations and quadratic equations. Is there a formula for the equations in the form $ax^4+bx^3+cx^2+dx+f=0$ I also skipped $e$ because of Euler's constant.
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2It takes all of 2 seconds to google it. – dxiv Mar 06 '18 at 04:06
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Here is my version:
Let the biquadratic be $$a_0 x^4 +4a_1 x^3 +6a_2 x^2 +4a_3 x + a_4=0$$ and the two invariants are
$$g_2=a_0 a_4 -4 a_1 a_3 +3 a_2^2$$ $$g_3=a_0 a_2 a_4 -a_0 a_3^2 -a_1^2 a_4 +2 a_1a_2 a_3 -a_2^3.$$ Then the resolvent cubic is $$4z^3 -g_2 z -g_3=0$$
with roots
$$u_0=\frac{1}{2}\sqrt[3]{g_3 +\sqrt{g_3^2 +\left(\frac{g_2}{3}\right)^3}}+\frac{1}{2}\sqrt[3]{g_3 -\sqrt{g_3^2 +\left(\frac{g_2}{3}\right)^3}}$$ $$u_1= \frac{\zeta^2}{2}\sqrt[3]{g_3 +\sqrt{g_3^2 +\left(\frac{g_2}{3}\right)^3}}+\frac{\zeta}{2}\sqrt[3]{g_3 -\sqrt{g_3^2 +\left(\frac{g_2}{3}\right)^3}}$$ $$u_2=\frac{\zeta}{2}\sqrt[3]{g_3 +\sqrt{g_3^2 +\left(\frac{g_2}{3}\right)^3}}+\frac{\zeta^2}{2}\sqrt[3]{g_3 -\sqrt{g_3^2 +\left(\frac{g_2}{3}\right)^3}}$$
And the roots of the biquadratic are given by
$$r_0=-\frac{1}{a_0} \left[a_1 + \sqrt{a_1^2 -a_0a _2 -a_0 u_0 } + \sqrt{a_1^2 -a_0a _2 -a_0 u_1 } + \sqrt{a_1^2 -a_0a _2 -a_0 u_2 }\right]$$ $$r_1=-\frac{1}{a_0}\left[a_1 + \sqrt{a_1^2 -a_0a _2 -a_0 u_0 } - \sqrt{a_1^2 -a_0a _2 -a_0 u_1 } - \sqrt{a_1^2 -a_0a _2 -a_0 u_2 }\right]$$ $$r_2=-\frac{1}{a_0}\left[a_1 - \sqrt{a_1^2 -a_0a _2 -a_0 u_0 } - \sqrt{a_1^2 -a_0a _2 -a_0 u_1 } + \sqrt{a_1^2 -a_0a _2 -a_0 u_2 }\right]$$ $$r_3=-\frac{1}{a_0}\left[a_1 -\sqrt{a_1^2 -a_0a _2 -a_0 u_0 } + \sqrt{a_1^2 -a_0a _2 -a_0 u_1 } - \sqrt{a_1^2 -a_0a _2 -a_0 u_2 }\right]$$
Rene Schipperus
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