Exact what a wrote below. The lady made a false conclusion. Only x = 0.5 + and - i(0.75)^0.5 is right for both terms and to get -2 for second term. x = -1 works only for the second one with exponent 2025 to geht -2.
@@n_kachi Well x^3 is equal to -1, so x = cube ruut of -1. If we put that number (cube ruut of -1) into the original equation, then I think the answer is one.
ye youre right but he never said that in the video .... x^3 = -1 is the solution to the eqn ... x^3 = -1 has three roots: (-1) and 2 complex roots which are the solution to the original equation ... and also he never put x=(-1) to find the answer but instead put x^3 = (-1)
First of all x +1/x =1 is not possible for any real value of x as A.M > G.M so x+1/x is always greater than or equal to 2 and the roots for this equation would be non real
This is a dangerous exercise if and when the concept of complex numbers is not mentioned. The way the presenter shows is just a symbolic magic, which is not true math teaching. When the equation x^3 = -1 is shown or the original quadratic equation is solved, we must stop and ask if there is a solution, and if yes, in what sense? The solution makes sense only in the complex domain.
Its all wrong You never should multiply to anything that contains variable in equation if it can be equal 0 You just multiplied to (x+1) and have got additional root x=1 So you can multiply to anything (x-2) (x-pi) (x+e)… And received infinite number of roots that not exists. Because of this you should never do that
@@kunalmachra my friend it means x=-1 and two other complex numbers which do exist in conjugate pair , u cant calculate some value of variable that doesnt even satisfy the original given equation , my friend maybe you should stick to working on real numbers first then shift to complex
If you use Wolfram Alpha to calculate x^2025 using the two complex x values you calculated, you will see that both equal -1. Therefore, the solution is correct!
@@ioannismichalopoulos6936I know that, but x= -1 is not a solution for the first equation. Only the complex solution I wrote above works for both and to get -2 for the second term with exponent 2025. Ergo, the lady made a false conclusion. 🤷
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-1 is not even a solution to the original equation
Exact what a wrote below. The lady made a false conclusion. Only x = 0.5 + and - i(0.75)^0.5 is right for both terms and to get -2 for second term. x = -1 works only for the second one with exponent 2025 to geht -2.
@@ganymed1236 That should be x = 0.5 + *or* - i(0.75)^0.5.
@@n_kachi Well x^3 is equal to -1, so x = cube ruut of -1. If we put that number (cube ruut of -1) into the original equation, then I think the answer is one.
ye youre right but he never said that in the video .... x^3 = -1 is the solution to the eqn ... x^3 = -1 has three roots: (-1) and 2 complex roots which are the solution to the original equation ... and also he never put x=(-1) to find the answer but instead put x^3 = (-1)
The only different approach I would use is:
x^2+1=x
x^2=x-1
x^3=x^2-x=x-1-x=-1
Took a glance at it, and.. yeah, there's no solution
(D= b^2 - 4ac = 1-4 =-3)
First of all x +1/x =1 is not possible for any real value of x as A.M > G.M so x+1/x is always greater than or equal to 2 and the roots for this equation would be non real
This is a dangerous exercise if and when the concept of complex numbers is not mentioned. The way the presenter shows is just a symbolic magic, which is not true math teaching. When the equation x^3 = -1 is shown or the original quadratic equation is solved, we must stop and ask if there is a solution, and if yes, in what sense? The solution makes sense only in the complex domain.
Is that true?
Посчитал формулой пика за -1 секунду
(a-b)²=(a-b)(a-b)=a²-ab-ab+b²=a²-2ab+b²
Hope have helped
Its all wrong
You never should multiply to anything that contains variable in equation if it can be equal 0
You just multiplied to (x+1) and have got additional root x=1
So you can multiply to anything
(x-2)
(x-pi)
(x+e)…
And received infinite number of roots that not exists.
Because of this you should never do that
Ótimo
x = -1 doesnt even satsfy the given q , utter bs
X^3=-1 does not mean X=-1, ever heard of complex numbers?
@@kunalmachra my friend it means x=-1 and two other complex numbers which do exist in conjugate pair , u cant calculate some value of variable that doesnt even satisfy the original given equation , my friend maybe you should stick to working on real numbers first then shift to complex
That is nonsens. Complex solutions for x+1/x=1 are 0,5 + and - i(0,75)^0,5.
If you use Wolfram Alpha to calculate x^2025 using the two complex x values you calculated, you will see that both equal -1. Therefore, the solution is correct!
Make sure though that you use . instead of , as a decimal point
@@ganymed1236 x = (1+i√3)/2
using (a+b)^3 formula
x^3 = (1+3√3 i+3(- 3)+3√3 i^3)/8
= (1 - 9)/8 = - 1 because i^2 = - 1 & i^3 = - i
similarly if x = (1 - i√3)/2 then x^3 = - 1
@@ioannismichalopoulos6936I know that, but x= -1 is not a solution for the first equation. Only the complex solution I wrote above works for both and to get -2 for the second term with exponent 2025. Ergo, the lady made a false conclusion. 🤷
Original poster, you are to use "or," not "and." x = 0.5 + or - i(0.75)^0.5
que dublagem ruim