Can We Solve A Nice Factorial Equation

🤩 Hello everyone, I'm very excited to bring you a new channel (aplusbi)
Enjoy...and thank you for your support!!! 🧡🥰🎉🥳🧡

If you need to post a picture of your solution or idea:
#radicals #radicalequations #algebra #calculus #differentialequations #polynomials #prealgebra #polynomialequations #numbertheory #diophantineequations #comparingnumbers #trigonometry #trigonometricequations #complexnumbers #math #mathcompetition #olympiad #matholympiad #mathematics #sybermath #aplusbi #shortsofsyber #iit #iitjee #iitjeepreparation #iitjeemaths #exponentialequations #exponents #exponential #exponent #systemsofequations #systems
#functionalequations #functions #function #maths #counting #sequencesandseries #sequence #factorials #factorial #factorialequations
via @YouTube @Apple @Desmos @NotabilityApp @googledocs @canva

SIMILAR PROBLEM

PLAYLISTS 🎵 :

[aweebthatlovesmath]

1:07 i think x!=x^3-x or gamma(x+1)=x^3-x has infinitely many solutions over complex numbers and since they can't be nicely parameterized it just shows some of it but you can write your equation and then type your set of solutions to be from what sets and it gives you them!!

[davidtaran]

1. check that x=0 not a root. divide by x. get:
(x-1)! = x^2 - 1 = (x-1)(x+1).
2. check that x=1 not a root. divide by (x-1). get:
(x-2)! = x+1
3. from intersection graphs of functions y=(x-2)! and y=x+1 and x>1 get:
x=5

[ФролПрохацкий]

Я ограничился тем, что нашёл корень подбором и предположил что он один, примерно понимая что на графике, если речь о натуральных X

[allanmarder]

x!= x^3 -x so x!= x*(x-1)*(x+1) or x*(x-1)*(x-2)! = x*(x-1)*(x+1) [x*(x-1)]*[(x-2)! -(x+1)] Now 0 or 1 won't work, so we must have
(x-2)! = x+1 At this point I reasoned as follows: x+1 must have an integer factorial value so check 1, 2, 6, 24 etc. Now
1.and 2 won't work. Also 24 is clearly too large. The only one that has a chance is 6 which leads to x=5 and (5-2)!=6.

[FisicTrapella]

(x-2)! = (x+1)
The equality must be (x-2)(x-3) = (x+1) in order to have an x in the LHS
So, we get x^2 - 6x + 5 = 0 with solutions 1 and 5. As we know x=1 can't be a solution and x=5 is the only integer solution. 🤗

[govin]

#solution be non interger not.possibls but are 2.61803399 and2.61803399 by quadratic 😢