Solving A Nice Equation

🤩 Hello everyone, I'm very excited to bring you a new channel (SyberMath Shorts).
Enjoy...and thank you for your support!!! 🧡🥰🎉🥳🧡
/ @sybermath
/ @aplusbi
⭐ Join this channel to get access to perks:→ bit.ly/3cBgfR1
My merch → teespring.com/...
Follow me → / sybermath
Subscribe → www.youtube.co...
⭐ Suggest → forms.gle/A5bG...
If you need to post a picture of your solution or idea:
in...
#algebra #polynomials #polynomial
via @RUclips @Apple @Desmos @NotabilityApp @googledocs @canva
PLAYLISTS 🎵 :
▶ Trigonometry: • Trigonometry
▶ Algebra: • Algebra
▶ Complex Numbers: • Complex Numbers
▶ Calculus: • Calculus
▶ Geometry: • Geometry
▶ Sequences And Series: • Sequences And Series

[@awahan]

You can also treat as a difference of two squares, then you can factor it directly

[@DarkLantern1]

You can also divide both sides by (x-1)^4 and treat it as a 4th root of unity question

[@Blaqjaqshellaq]

In addition, x^4 - (x-1)^4=[x^2+(x-1)^2]*[x^2-(x-1)^2]=[x^2+(x-1)^2]*[x+(x-1)]*[x-(x-1)].
If x^4 - (x-1)^4=0, it follows that (2*x^2-2*x+1)*(2*x-1)*2=0.
If 2*x - 1=0, then x=1/2, giving us one of the roots.
If 2*x^2 - 2*x +1=0, this quadratic equation gives us the other two roots, (1+i)/2 and (1-i)/2.

[@r4_in_space]

When a function is shifted, that's called a translation.

[@Gezraf]

there's a method similar to the 2nd method. instead of relying on conjugates, you could divide both sides by (x-1)^4, so that:
(x/(x-1))^4 = 1
take the 4th root
x/(x-1) = +-1
x/(x-1) = 1 --> x = x-1 --> false
x/(x-1) = -1 --> x = 1 - x --> 2x = 1 --> x = 1/2

[@williamperez-hernandez3968]

For 2nd Method, instead of taking 4th root, take only square root. Then we get two equations: 1. x^2 = (x-1)^2, and 2. x^2 = - (x-1)^2. Equation 1 simplifies to 2x=1, so x=1/2. Equation 2 simplifies to 2x^2 - 2x +1 = 0, with complex solutions x = (1+ i)/2 and x = (1-i)/2.

[@ytlongbeach]

Yes, this is exactly what i did, also !!

[@elmer6123]

Substitute x=y+1/2 into the given equation and see how many terms cancel out.
(y+1/2)^4-(y-1/2)^4=2*4[(1/2)y^3+(1/2)^3*y]=0
Factor out 4y: 4y(y^2+1/4)=0, so y=0 or y=±i/2
Thus, x=y+1/2=1/2 or (1±i)/2

[@trojanleo123]

You used variable "u" and forgot to say Happy birthday to U. 🤣🤣🤣

[@phill3986]

👍👏😀☮️✌️✌️☮️😀👏👍

[@key_board_x]

x⁴ = (x - 1)⁴
x⁴ - (x - 1)⁴ = 0
(x²)² - [(x - 1)²]² = 0 → recall: a² - b² = (a + b).(a - b)
[x² + (x - 1)²].[x² - (x - 1)²] = 0
[x² + (x² - 2x + 1)].[x² - (x² - 2x + 1)] = 0
[x² + x² - 2x + 1].[x² - x² + 2x - 1] = 0
(2x² - 2x + 1).(2x - 1) = 0
First case: (2x + 1) = 0
2x + 1 = 0
2x = - 1
x = - 1/2
Second case: (2x² - 2x + 1) = 0
2x² - 2x + 1 = 0
Δ = (- 2)² - 4.(2 * 1) = 4 - 8 = - 4 = 4i² = (2i)²
x = (2 ± 2i)/4
x = (1 ± i)/2

[@prollysine]

by faktoring , multp. *(-1/2) , 4x^3 -- 6x^2 + 4x - 1 = 0 , (2x-1)(2x^2-2x+1)=0 , 2x-1=0 , x=1/2 ,
+4 -2 2x^2-2x+1=0 , x=(2+/-V(4-8))/4 , x=(2+/-V(-4))/4 ,
-4 +2 x=(2+/-i*V(4))/4 , x= (1/2+i/2) , (1/2-i/2) ,
+2 -1 = 0 , test , (1/2)^4=1/16 , (1/2-1)^4=1/16 ,
(1/2+i/2)^4=-1/4 , (1/2+i/2-1)^4=-1/4 , OK ,
solu , x= 1/2 , (1/2+i/2) , (1/2-i/2) , (1/2-i/2)^4=-1/4 , (1/2-i/2-1)^4=-1/4 , OK ,

[@Don-Ensley]

x ∈ { ½, ½ - ⁱ/₂, ½ + ⁱ/₂ }