Solving A Nice Cubic Equation

[@nasrullahhusnan2289]

x³+x=⅝ --> x(x²+1)=⅛(4+1) =½×¼(4+1) =½(1+¼ ) =½[1+(½)²] Thus x=½ 0:13

[@stvp68]

My motto is no pain, no pain 😅

[@robertibatullin8329]

Simplification of x in the 2nd method is a problem in its own right.

[@LazarAndric-se4cq]

7:30 you missed the sign of 1/27

[@meunomeeeu3118]

No, he didnt

[@raghvendrasingh1289]

👍 8 x^3+8 x - 5 = 0 there is only one sign change in f(x) and no sign change in f(-x) hence by Descartes' rule of signs equation has only one real root and it is positive. by RRT we obtain root , x = 1/2

[@Don-Ensley]

problem x³ + x = 5/8 Cubic formula x = a + b a³ + b³ = 5/8 -3ab = 1 ab = -1/3 a³ b ³ = -1/27 b³ = -1/(27 a³) a³ -1/(27 a³) = 5/8 27 (a³)²-(5/8)27 (a³)-1 = 0 a³ = { (5/8)27 ± √[(5/8)² 27² +4(27)] }/54 Δ = (5/8)² 27² +4(27) = (27)(25•27/64 +256/64) = 9•3/(64) (25•27+256) = 9•3/(64) 931 a³ = { (5/8)27 ± (3/8)√[3•931] }/54 = { 135 ± 21√57} / 432 = { 45 ± 7√57} / 144 x = a + b = ∛ [ (45 + 7√57) / 144 ] + ∛ [ (45 - 7√57) / 144 ] = 1/2 This is one real root. x -1/2 a factor. x³ + x - 5/8 = 0 (x -1/2) x² +(x-1/2) x/2+(5/4)(x-1/2) = 0 (x-1/2)(x² +x/2 +5/4) = 0 ZPP and Quadratic x = (-1/2 ± √[(1/4)-5]}/2 = -1/4 ± 1/4 i √19 = (-1 ± i √19 )/4 answer x ∈ { 1/2, (-1 - i √19 )/4, (-1 + i √19 )/4 }

[@VittorioBalbi1962]

😂