An Interesting Quartic Equation

[@اسماعیلخسروی-خ6ظ]

Excellent ❤❤❤❤❤❤❤

[@Don-Ensley]

problem
x⁴ + 12x + 3 = 0
Using unknown coefficients a, b and c factor into a product of 2 quadratics.
(x² + ax + b) (x²- ax + c) = 0
Expand to generate a coefficient system of equations.
x⁴ (c - a² + b) x² + (ac - ab) x + bc = 0
c - a² + b = 0
ac - ab = 12
bc = 3
c + b = a²
c - b = 12/a
bc = 3
c = (a² + 12/a) / 2
b = (a² - 12/a) / 2
(a² + 12/a)(a² - 12/a) / 4 = 3
a⁴-144/( a² ) = 12
( a² )³ - 12 ( a²) - 144 = 0
Let
y = a²
y³ - 12 y - 144 = 0
Use the rational root theorem (RRT).
216 - 72 - 144 = 0
For the root
y = 6
= a²
System solution using the positive root of a² :
a = √6
c = (3 + √6)
b = (3 - √6)
Factored quartic:
x⁴ + 12x + 3 =
(x² + √6 x + 3 -√6) (x²- √6 x + 3+√6) = 0
Apply quadratic formula twice.
From x² + √6 x + 3 -√6 = 0:
x ={ -√6 ± √ [6-4(3-√6)]}/2
Δ = 6-4(3-√6)
= 4√6 - 6)
x = [ -√6 ± √(4√6- 6) ] / 2
From x²- √6 x + 3+√6 = 0
x. ={ √6 ± √ [6-4(3+√6)]}/2
Δ = 6-4(3+√6)
=-(4√6 + 6)
x = [ √6 ± i √(4√6 + 6) ] / 2
x ∈ { [ -√6 - √(4√6 - 6) ] / 2,
[ -√6 + √(4√6 - 6) ] / 2,
[ √6 - i √(4√6 +6) ] / 2,
[ √6 + i √(4√6 +6) ] / 2 }

[@raghvendrasingh1289]

❤
Descartes' method
however from
y^3 - 12 y = 144
we can easily find the root y = 6 by RRT

[@Don-Ensley]

@@raghvendrasingh1289thanks! I have modified my comment to include your observation. ❤️🙏

[@scottcowan8036]

You posted this same video on your main channel on Feb 3, 2023.

[@yoav613]

Noice😊