I would not have accepted zero (0) as a root. When inserted into the •original• equation it leads into division by zero, which is definitely contraindicated in mathematics. Anyone?
For real numbers, this equation reduces to 3 = x - 1 + 1/x, and this reduces to a simple quadratic, x^2-4x+1=0. Zero is not a solution to the given equation, and there are no complex roots. The only solution are x = 2 + sqrt(3) or x= 2 - sqrt(3).
I dont believe the accuracy of the complex solutions. For example x=-0.5+sqrt(3)/2*i. cos(2pi/3)=-0.5 and sin(2pi/3)=sqrt(3)/2. Hence: x^2=cos(4pi/3)+isin(4pi/3)=-0.5-i*sqrt(3)/2 x^3=cos(2pi)+isin(2pi)=1 x^4=x=-0.5+i*sqrt(3)/2 x^5=x^2=-0.5-i*sqrt(3)/2 x^6=cos(4pi)+isin(4pi)=1 x^7=x=-0.5+i*sqrt(3)/2 x^4+x^5+x^6=0. x^3+x^5+x^7=0 Clearly 0/03 so this proposed solution fails. The other complex solution is x^2 in the above, i.e.leads to the same 0/0 , clearly not a solution.
Num: x^3*(x^4+x^2+1) =x^3*(x^6-1)/(x^2-1) as sum of series =(x^3)*(x^3+1)*(x^3-1)/(x-1)/(x+1) =(x^3)*(x^2-x+1)*(x^2+x+1) Den: (x^4)*(x^2+x+1) Hence fraction becomes: (x^2-x+1)/x. Equation now x^2-x+1=3x, so x^2-4x+1=0. Discrim=16-4=12 Solutions 2+sqrt(3),2-sqrt(3) Checked both solutions with calculator and ratio 3 is obtained.
That is an unexpected number theory adjacent problem. The solutions are x = 0, the two cube roots of unity, 2+/-sqrt(3). Also when I say two cube roots of units, I DO mean (-1+/-sqrt(3)i)/2. I will have to use that for both advanced algebra and refreshing my knowledge of number theory!!!
Х=0, 0⁷+0⁵+0³/ 0⁶+0⁵+0⁴ ≠3???? ≠0????? На 0 делить нельзя...It cannot be divided by 0... Thank you.😢 (Greetings from Russia, Veliky Novgorod. Best wishes.👍 Grandfather is 62 years old. A former high school student, and a student.😂 Sometimes I watch your videos....I'm listening to music. Interesting.)
I would not have accepted zero (0) as a root. When inserted into the •original• equation it leads into division by zero, which is definitely contraindicated in mathematics. Anyone?
You're definitely right
There's more (other commentators helped me to see this): x^2 + x + 1 ≠ 0.
So complex solutions don't fit either.
Not first determining the range of acceptable values is a grave mistake!
True, only irrational real solutions are acceptable
For real numbers, this equation reduces to 3 = x - 1 + 1/x, and this reduces to a simple quadratic, x^2-4x+1=0. Zero is not a solution to the given equation, and there are no complex roots. The only solution are x = 2 + sqrt(3) or x= 2 - sqrt(3).
I dont believe the accuracy of the complex solutions. For example x=-0.5+sqrt(3)/2*i.
cos(2pi/3)=-0.5 and sin(2pi/3)=sqrt(3)/2.
Hence:
x^2=cos(4pi/3)+isin(4pi/3)=-0.5-i*sqrt(3)/2
x^3=cos(2pi)+isin(2pi)=1
x^4=x=-0.5+i*sqrt(3)/2
x^5=x^2=-0.5-i*sqrt(3)/2
x^6=cos(4pi)+isin(4pi)=1
x^7=x=-0.5+i*sqrt(3)/2
x^4+x^5+x^6=0.
x^3+x^5+x^7=0
Clearly 0/03 so this proposed solution fails.
The other complex solution is x^2 in the above, i.e.leads to the same 0/0 , clearly not a solution.
Х=0?! Ничего не смутило?
Ну бывают люди с нулевыми знаниями в элементарной математике, а у автора ролика этих знаний в 3 раза больше!🤣🤣🤣
Num:
x^3*(x^4+x^2+1)
=x^3*(x^6-1)/(x^2-1) as sum of series
=(x^3)*(x^3+1)*(x^3-1)/(x-1)/(x+1)
=(x^3)*(x^2-x+1)*(x^2+x+1)
Den:
(x^4)*(x^2+x+1)
Hence fraction becomes:
(x^2-x+1)/x.
Equation now x^2-x+1=3x, so x^2-4x+1=0. Discrim=16-4=12
Solutions 2+sqrt(3),2-sqrt(3)
Checked both solutions with calculator and ratio 3 is obtained.
That is an unexpected number theory adjacent problem. The solutions are x = 0, the two cube roots of unity, 2+/-sqrt(3). Also when I say two cube roots of units, I DO mean (-1+/-sqrt(3)i)/2. I will have to use that for both advanced algebra and refreshing my knowledge of number theory!!!
Про х=0 було смішно.
Спасибо, что учите под музику тихую
Х=0, 0⁷+0⁵+0³/ 0⁶+0⁵+0⁴ ≠3???? ≠0????? На 0 делить нельзя...It cannot be divided by 0... Thank you.😢
(Greetings from Russia, Veliky Novgorod. Best wishes.👍 Grandfather is 62 years old. A former high school student, and a student.😂 Sometimes I watch your videos....I'm listening to music. Interesting.)