# In Search for Radical Roots of Polynomial Equations of degree n > 1

Take note: Find roots (根) to solve polynomial (多项式方程式) equations, but find solutions (解) to solve simultaneous equations (联式方程式).

Radical : (Latin Radix = root): $\sqrt [n]{x}$

Quadratic equation (二次方程式) [最早发现者 : Babylon  和 三国时期的吴国 数学家 赵爽]

${a.x^{2} + b.x + c = 0}$

$\boxed{x= \frac{-b \pm \sqrt{b^{2}-4ac} }{2a}}$

Cubic Equation: 16 CE Italians del Ferro,  Tartaglia & Cardano
${a.x^{3} = p.x + q }$

Cardano Formula (1545 《Ars Magna》):
$\boxed {x = \sqrt [3]{\frac {q}{2} + \sqrt{{ (\frac {q}{2})}^{2} - { (\frac {p}{3})}^{3}}} + \sqrt [3]{\frac {q}{2} -\sqrt{ { (\frac {q}{2})}^{2} - { (\frac {p}{3})}^{3}}}}$

Example:
${x^{3} = 15x + 4}$
By obvious guess,  x = 4
Using Cardano formula,
$x = \sqrt[3]{2+ 11 \sqrt{-1}} + \sqrt [3]{2 - 11 \sqrt{-1}}$

They discovered the first time in history the “Imaginary” number (aka Complex number):
$\boxed {i = \sqrt{-1}}$
then
$(2 + i)^{3} =2+11i$
$(2 - i)^{3} =2-11i$
$x = (2 + i) + (2 - i) = 4$

Quartic Equation: by Cardano’s student Ferrari
${a.x^{4} + b.x^{3} + c.x^{2} + d.x + e = 0}$

Quintic Equation:
${a.x^{5} + b.x^{4} + c.x^{3} + d.x^{2} + e.x + f = 0}$

No radical solution (Unsolvability) was suspected by Ruffini (1799), proved by Norwegian Abel (1826), but explained by French 19-year-old boy Évariste Galois (discovered in 1831, published only after his death in 1846) with his new invention : Abstract Algebra “Group“(群) & “Field” (域)。

Notes:

Field Theory is Elementary Math.

Field is the Algebraic structure which has 4 operations on calculator (+ – × ÷). Examples : Rational number $(\mathbb{Q})$, Real $(\mathbb{R})$, Complex $(\mathbb{C})$, $\mathbb{Z}_{p}$  (Integer modulo prime, eg.Z2 = {0, 1}) , etc.

If $\mathbb{Q}$   (“a”, “b”) is adjoined with irrational (eg. $\sqrt {2}$)  to become a larger Field (extension) $\mathbb{Q} (\sqrt {2}) = a +b\sqrt {2}$
it has a beautiful “Symmetry” aka Conjugate
$(a - b\sqrt {2})$

Field Extension of $\mathbb{Q} (\sqrt {2}) = a +b\sqrt {2}$ :

Any equation P(x) = 0
with root in $\mathbb{Q} (\sqrt {2}) = a +b\sqrt {2}$ will have
another conjugate root $(a - b\sqrt {2})$

Galois exploited such root symmetry in his Group structure to explain the unsolvability for polynomial equations of quintic degree and above.

Ref: 《Elements of Mathematics – From Euclid to Gödel》by John Stillwell (Princeton University Press, 2016) [NLB # 510.711]