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Polynomials with order 3 have at most 3 solutions. Arranging the eqn in another order: x^3 - 27 = 0 x^3 - 3^3 = 0 (x - 3) (x^2 + 3x +9) = 0 Apparently, x = 3 and there are 2 more complex solutions which can be obtained by quadratic formula. DONE😊
Use Euler's formula for questions like this. (x/3)^3 = 1 = e^(2πi)=> roots e^0, e^(2πi/3), e^(4π/3). Solutions 3 x the above list so 3, 3(-1 +√3 i)/2, 3(-1 - √3i)/2.
In my simple mind, the answeres would be 3, and absolute value of 3 and -3, the remainder of the solutions seem to be changing the equation. Someone who is smarter than I am, please explain why you would care about creating a negative... this seems overly complicated for the sake of complicating it.
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Polynomials with order 3 have at most 3 solutions.
Arranging the eqn in another order:
x^3 - 27 = 0
x^3 - 3^3 = 0
(x - 3) (x^2 + 3x +9) = 0
Apparently, x = 3 and there are 2 more complex solutions which can be obtained by quadratic formula.
DONE😊
Perfect math channel, but please put the dividing line (supra, over) next to the equal sign, that's how it's done :)
just 3 seconds 😊
Use Euler's formula for questions like this. (x/3)^3 = 1 = e^(2πi)=> roots e^0, e^(2πi/3), e^(4π/3).
Solutions 3 x the above list so 3, 3(-1 +√3 i)/2, 3(-1 - √3i)/2.
Only one cube root is 3...
In my simple mind, the answeres would be 3, and absolute value of 3 and -3, the remainder of the solutions seem to be changing the equation. Someone who is smarter than I am, please explain why you would care about creating a negative... this seems overly complicated for the sake of complicating it.
All solutions are 3, 3*e^(2/3pi* i) and 3 *(e^( 4/3pi* i)
Simply don't waste your time & viewer's time by writing many steps unnecessarily.
X=3
Help (x+y)^x^1/2=(x-y)^y^1/2
2 х 2 = Х.
Х = ?
Read the question. How many roots. The answer is 3. And you want to waste our time. Go attend English classes
X=3