一次方程的求根公式
\[x = {-b \over a}
\]
The linear formula gives the solution of \(ax+b=0\) for real numbers \(a\), \(b\) with \(a\neq0\).
二次方程的求根公式
\[x = {-b\pm\sqrt{b^2-4ac} \over 2a}
\]
The quadratic formula gives the solutions of \(ax^2+bx+c=0\) for real numbers \(a\), \(b\), \(c\) with \(a\neq0\).
三次方程的求根公式
\[x = {-2b + \bigl({-1+\sqrt{-3}\over2}\bigr)^n\sqrt[3]{4\bigl(-2b^3+9abc-27a^2d+\sqrt{(-2b^3+9abc-27a^2d)^2-4(b^2-3ac)^3}\bigr)}+\bigl({-1-\sqrt{-3}\over2}\bigr)^n\sqrt{3}{4\bigl(-2b^3+9abc-27a^2d-\sqrt{(-2b^3+9abc-27a^2d)^2-4(b^2-3ac)^3}\bigr)} \over 6a}
\]
The cubic formula gives the solutions of \(ax^3+bx^2+cx+d=0\) for real numbers \(a\), \(b\), \(c\), \(d\) with \(a\neq0\).
Directions: Take \(n=0\), \(1\), \(2\). Use real cube roots if possible, and principal roots otherwise.
四次方程的求根公式
\[\def\sgn{\mathop{\rm sgn}} x = {-3b\pm\biggl(\sqrt{3\Bigl(3b^2-8ac+2a\sqrt[3]{4\bigl(2c^3-9bcd+27ad^2+27b^2e-72ace+\sqrt{(2c^3-9bcd+27ad^2+27b^2e-72ace)^2-4(c^2-3bd+12ae)^3}\bigr)}+2a\sqrt[3]{4\bigl(2c^3-9bcd+27ad^2+27b^2e-72ace-\sqrt{(2c^3-9bcd+27ad^2+27b^2e-72ace)^2-4(c^2-3bd+12ae)^3}\bigr)}\Bigr)}\pm\sqrt{3\Bigl(3b^2-8ac+2a\bigl({-1+\sqrt{-3}\over2}\bigr)\sqrt[3]{4\bigl(2c^3-9bcd+27ad^2+27b^2e-72ace+\sqrt{(2c^3-9bcd+27ad^2+27b^2e-72ace)^2-4(c^2-3bd+12ae)^3}\bigr)}+2a\bigl({-1-\sqrt{-3}\over2}\bigr)\sqrt[3]{4\bigl(2c^3-9bcd+27ad^2+27b^2e-72ace-\sqrt{(2c^3-9bcd+27ad^2+27b^2e-72ace)^2-4(c^2-3bd+12ae)^3}\bigr)}\Bigr)}\biggr)\pm\sgn\biggl(\Bigl(\sgn\bigl(-b^3+4abc-8a^2d\bigr)-{1\over2}\Bigr)\Bigl(\sgn\bigl(\max((2c^3-9bcd+27ad^2+27b^2e-72ace)^2-4(c^2-3bd+12ae)^3,\min(3b^2-8ac,3b^4+16a^2c^2+16a^2bd-16ab^2c-64a^3e))\bigr)-{1\over2}\Bigr)\biggr)\sqrt{3\Bigl(3b^2-8ac+2a\bigl({-1-\sqrt{-3}\over2}\bigr)\sqrt[3]{4\bigl(2c^3-9bcd+27ad^2+27b^2e-72ace+\sqrt{(2c^3-9bcd+27ad^2+27b^2e-72ace)^2-4(c^2-3bd+12ae)^3}\bigr)}+2a\bigl({-1+\sqrt{-3}\over2}\bigr)\sqrt[3]{4\bigl(2c^3-9bcd+27ad^2+27b^2e-72ace-\sqrt{(2c^3-9bcd+27ad^2+27b^2e-72ace)^2-4(c^2-3bd+12ae)^3}\bigr)}\Bigr)}\over12a}
\]
The quartic formula gives the solutions of \(ax^4+bx^3+cx^2+dx+e=0\) for real numbers \(a\), \(b\), \(c\), \(d\), \(e\) with \(a\neq0\).
Directions: Choose all possibilities for the three \(\pm\) signs with the last two equivalent. Use real cube roots if possible, and principal roots otherwise.