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!!
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
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.
My route to glory -> x!=x^3-x=(x+1)x(x-1). If x! has 2 terms, then x+1=1, not a solution, for x=0. If x! has 3 terms then x(x-1)(x-2)=(x+1)x(x-1) then x-2=x+1, so no solution. If x! has 4 terms then x(x-1)(x-2)(x-3)=(x+1)x(x-1) then (x-2)(x-3)=x+1 or x^2-5x+6=x+1 or x^2-6x+5=0. Factoring this gives, (x-5)((x-1)=0 but x > 1, so x=5. 5!=120=125-5 so checks out.
(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. 🤗
I got x=5 by guessing and checking.
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!!
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
Я ограничился тем, что нашёл корень подбором и предположил что он один, примерно понимая что на графике, если речь о натуральных X
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.
My route to glory -> x!=x^3-x=(x+1)x(x-1). If x! has 2 terms, then x+1=1, not a solution, for x=0. If x! has 3 terms then x(x-1)(x-2)=(x+1)x(x-1) then x-2=x+1, so no solution. If x! has 4 terms then x(x-1)(x-2)(x-3)=(x+1)x(x-1) then (x-2)(x-3)=x+1 or x^2-5x+6=x+1 or x^2-6x+5=0. Factoring this gives, (x-5)((x-1)=0 but x > 1, so x=5. 5!=120=125-5 so checks out.
(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. 🤗
But guess 5 rigyt😂
0
x = 5
i made this via linkedin
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@@SyberMath this is an example i found via my linkedin account
#solution be non interger not.possibls but are 2.61803399 and2.61803399 by quadratic 😢