@@domedebali632 Yes brother In this equation It has two real solution and one complex solution However i provide in exact form x=2[Real solution] x=-(1/2)-((√5)/2))[Real solution] x=-(1/3)-(((∛49)(-2(ω)))/(∛((36)+i(108√3))))+(ω/3)(∛((7/2)+i((21√3)/(2))))[complex solution]
@@domedebali632 Yes brother In this equations It has Real solution-2 Complex solution-1 Here i proceed right now x=2[Real solution] x=-(1/2)-(√5/2)[Real solution] x=-(1/3)-(∛(392)ω)/(∛(1323+i(√(47258883)))+((∛((7+i(√1323))/(54)))ω)[complex solution]
@@domedebali632 @ Yes brother In this equations It has Real solution-2 Complex solution-1 Here i proceed right now x=2[Real solution] x=-(1/2)-(√5/2)[Real solution] x=-(1/3)-(∛(392)ω)/(∛(1323+i(√(47258883)))+((∛((7+i(√1323))/(54)))ω)[complex solution]
You already did this problem in May 2024, but for reasons unknown to me you have deleted your previous video. Having watched both your video and that of Prime Newtons from May 2024 on solving the equation (1) x³ − 3x = √(x + 2) I would like to make some additional observations because there is a lot more to say about this equation that neither you nor Prime Newtons discussed. Squaring both sides we obtain (2) x⁶ − 6x⁴ + 9x² − x − 2 = 0 This is a 6th degree equation which of course also has the roots of (3) x³ − 3x = −√(x + 2) since this gives (2) as well when both sides are squared. For anyone who is familiar with Chebyshev polynomials it is quite clear what we have here. If we substitute x = 2t in (2) we obtain (4) 64t⁶ − 96t⁴ + 36t² − 2t − 2 = 0 which can be written as (5) 32t⁶ − 48t⁴ + 18t² − 1 = t where the left hand side is a Chebyshev polynomial of the first kind. In fact the left hand side of (5) is T₆(t) and the right hand side is T₁(t) so we can rewrite (5) as (6) T₆(t) = T₁(t) Chebyshev polynomials of the first kind can be defined by the recurrence (7) T₀(x) = 1, T₁(x) = x, Tₙ₊₁(x) = 2x·Tₙ(x) − Tₙ₋₁(x) which means that they satisfy (8) Tₙ(cos θ) = cos nθ Therefore, if we substitute t = cos α in (5) or x = 2t = 2·cos α in (2) these equations reduce to (9) cos 6α = cos α and this trigonometric equation is easily solved. We have 6α = α + 2kπ ⋁ 6α = −α + 2kπ, k ∈ ℤ 5α = 2kπ ⋁ 7α = 2kπ, k ∈ ℤ α = ²⁄₅·kπ ⋁ α = ²⁄₇·kπ, k ∈ ℤ so for (2) we have the roots x₁ = 2·cos 0, x₂ = 2·cos(²⁄₅·π), x₃ = 2·cos(⁴⁄₅·π), x₄ = 2·cos(²⁄₇·π), x₅ = 2·cos(⁴⁄₇·π), x₆ = 2·cos(⁶⁄₇·π) Here x₁ = 2·cos 0 = 2 so the left hand side of (2) has a factor x − 2 and x₂ = 2·cos(²⁄₅·π) = −¹⁄₂ + ¹⁄₂√5, x₃ = 2·cos(⁴⁄₅·π) = −¹⁄₂ − ¹⁄₂√5 are the zeros of the quadratic polynomial x² + x − 1 which therefore also is a factor of the left hand side of (2). The remaining roots x₄ = 2·cos(²⁄₇·π), x₅ = 2·cos(⁴⁄₇·π), x₆ = 2·cos(⁶⁄₇·π) are the roots of the cubic equation (10) x³ + x² − 2x − 1 = 0 To prove this, we can substitute x = u + 1/u in (10) which results in (11) u⁶ + u⁵ + u⁴ + u³ + u² + u + 1 = 0 Since the left hand side of (11) is a geometric series with sum (u⁷ − 1)/(u − 1) we can also write this as (12) (u⁷ − 1)/(u − 1) = 0 Clearly, the roots of (12) and therefore of (11) are the seventh roots of unity except for u = 1 itself, so we have u = exp(k·²⁄₇·π·i), k = 1..6 and since x = u + 1/u this means that the roots of (10) are x = exp(k·²⁄₇·π·i) + exp(−k·²⁄₇·π·i) = 2·cos(k·²⁄₇·π), k = 1..6. Of course, for k = 4..6 we get the same values as for k = 1..3 so we indeed have the three roots x₄ = 2·cos(²⁄₇·π), x₅ = 2·cos(⁴⁄₇·π), x₆ = 2·cos(⁶⁄₇·π) for (10) and the left hand side of (10) is therefore also a factor of the left hand side of (2). WolframAlpha is incapable of coming up with the solutions x₄ = 2·cos(²⁄₇·π), x₅ = 2·cos(⁴⁄₇·π), x₆ = 2·cos(⁶⁄₇·π) for (10) and instead produces expressions for the solutions of (10) that contain cube roots of complex numbers even though all solutions of (10) are real. This is because the solutions of a cubic equation that has three distinct real but irrational roots cannot be expressed algebraically in radicals without using complex numbers. This is known as the _casus irreducibilis_ but there is a standard method to solve cubic equations with three distinct real but irrational roots trigonometrically, without using complex numbers. However, the intriguing thing here is that the expressions x₄ = 2·cos(²⁄₇·π), x₅ = 2·cos(⁴⁄₇·π), x₆ = 2·cos(⁶⁄₇·π) for the roots of (10) cannot be obtained either by using the standard trigonometric method to solve a cubic equation with three real roots. Substituting x = (z − 1)/3 in (10) results in (13) z³ − 21z − 7 = 0 and solving this depressed cubic trigonometrially in the conventional way we can express the roots of this equation as z₁ = 2√7·cos(¹⁄₃·arccos(¹⁄₁₄√7)) z₂ = 2√7·cos(¹⁄₃·arccos(¹⁄₁₄√7) − ²⁄₃·π) z₃ = 2√7·cos(¹⁄₃·arccos(¹⁄₁₄√7) + ²⁄₃·π) So, since x = (z − 1)/3, we can express the roots of (10) as x₄ = −¹⁄₃ + ²⁄₃√7·cos(¹⁄₃·arccos(¹⁄₁₄√7)) x₅ = −¹⁄₃ + ²⁄₃√7·cos(¹⁄₃·arccos(¹⁄₁₄√7) − ²⁄₃·π) x₆ = −¹⁄₃ + ²⁄₃√7·cos(¹⁄₃·arccos(¹⁄₁₄√7) + ²⁄₃·π) but it is not evident at all how these expressions can be reduced to the much simpler expressions x₄ = 2·cos(²⁄₇·π) x₅ = 2·cos(⁴⁄₇·π) x₆ = 2·cos(⁶⁄₇·π) Since we have proved that x − 2, x² + x − 1 and x³ + x² − 2x − 1 are all factors of the left hand side of (2) and since there are no other factors since any root of (2) is a zero of one of these factors it follows that we can write (2) as (14) (x − 2)(x² + x − 1)(x³ + x² − 2x − 1) = 0 Since the set of roots of (2) consist of the union of the set of roots of (1) and the set of roots of (3) we still need to find out which roots of (2) are roots of (1) and which roots of (2) are roots of (3). To do this, we can apply the double angle identity for the cosine to (9) which gives (15) 2·cos² 3α − 1 = 2·cos² ½α − 1 so we have cos² 3α = cos² ½α and therefore (9) is equivalent with (16) cos 3α = cos ½α ⋁ cos 3α = −cos ½α where the first equation is also obtained by substituting x = 2·cos α in (1) and the second by substituting this in (3) subject to the condition 0 ≤ α ≤ π to ensure the proper sign of √(x + 2) = cos ½α. Solving cos 3α = cos ½α gives α = ⁴⁄₅·kπ ⋁ α = ⁴⁄₇·kπ, k ∈ ℤ so taking the condition 0 ≤ α ≤ π into account we have the roots x₁ = 2·cos 0 = 2, x₃ = 2·cos(⁴⁄₅·π) = −¹⁄₂ − ¹⁄₂√5, x₅ = 2·cos(⁴⁄₇·π) for (1). Similarly, solving cos 3α = −cos ½α gives α = ²⁄₇·(2k + 1)π ⋁ α = ²⁄₅·(2k − 1)π, k ∈ ℤ and taking the condition 0 ≤ α ≤ π into account this gives the roots x₂ = 2·cos(²⁄₅·π) = −¹⁄₂ + ¹⁄₂√5, x₄ = 2·cos(²⁄₇·π), x₆ = 2·cos(⁶⁄₇·π) for (3).
I really appreciate the fact of using trigonometric functions in solving an algebraic problem
Thank you so much my friend for your support👍👍👍
Very nice video. I love math youtubers like you showing their faces and working on a problem nicely on the board
Haha thanks a lot my friend for your support👍👍👍
This is so nice. I really like your trig sub method
Thanks a lot my friend haha👍👍👍
x=2 is a solution
And we have two more solutions, and I believe that is the most tricky part on this problem bro
@@domedebali632
Yes brother
In this equation
It has two real solution and one complex solution
However i provide in exact form
x=2[Real solution]
x=-(1/2)-((√5)/2))[Real solution]
x=-(1/3)-(((∛49)(-2(ω)))/(∛((36)+i(108√3))))+(ω/3)(∛((7/2)+i((21√3)/(2))))[complex solution]
@@domedebali632
Yes brother
In this equations
It has
Real solution-2
Complex solution-1
Here i proceed right now
x=2[Real solution]
x=-(1/2)-(√5/2)[Real solution]
x=-(1/3)-(∛(392)ω)/(∛(1323+i(√(47258883)))+((∛((7+i(√1323))/(54)))ω)[complex solution]
@@domedebali632
@
Yes brother
In this equations
It has
Real solution-2
Complex solution-1
Here i proceed right now
x=2[Real solution]
x=-(1/2)-(√5/2)[Real solution]
x=-(1/3)-(∛(392)ω)/(∛(1323+i(√(47258883)))+((∛((7+i(√1323))/(54)))ω)[complex solution]
Another great video professor
Thank you so much my friend👍👍👍
This is so clever method prof🎉
Thank you my friend for your support👍👍👍
You are the best
Haha salute my friend👍👍👍
The only solution is x = 2. Plug the other values into a calculator to see that they do not solve the equation.
Nope...
Miklos shweitzers please
Yes for sure! Let me look into it haha👍👍👍
How cosin involved in answering the questions. I'm cooked
It is a trig substitution
dr pk math pls help me go from the continued fraction of the cube root of 2 back to the cube root of 2 using algebra I am need help
I am no dr pk but I did that problem a few months ago. I used infinite radical problems and got hints from there
@@Min-cv7nt did you solve it? I have a couple ideas but I dont like how it works
@@jjeastside Yes, I did. My solution was similar with infinite nested radicals
@@Min-cv7nt so if I gave you an infinite list of continued numbers could you tell me what cubic root it is?
@@Min-cv7nt I see it now instead of a single radical expanding its infinite radicals expanding in the form of
x = 1 + 1/(x^2 +x + 1)
Am I literally first ?
you beat me to it bro
Haha thanks for your support my friend👍👍👍
You already did this problem in May 2024, but for reasons unknown to me you have deleted your previous video. Having watched both your video and that of Prime Newtons from May 2024 on solving the equation
(1) x³ − 3x = √(x + 2)
I would like to make some additional observations because there is a lot more to say about this equation that neither you nor Prime Newtons discussed. Squaring both sides we obtain
(2) x⁶ − 6x⁴ + 9x² − x − 2 = 0
This is a 6th degree equation which of course also has the roots of
(3) x³ − 3x = −√(x + 2)
since this gives (2) as well when both sides are squared. For anyone who is familiar with Chebyshev polynomials it is quite clear what we have here. If we substitute x = 2t in (2) we obtain
(4) 64t⁶ − 96t⁴ + 36t² − 2t − 2 = 0
which can be written as
(5) 32t⁶ − 48t⁴ + 18t² − 1 = t
where the left hand side is a Chebyshev polynomial of the first kind. In fact the left hand side of (5) is T₆(t) and the right hand side is T₁(t) so we can rewrite (5) as
(6) T₆(t) = T₁(t)
Chebyshev polynomials of the first kind can be defined by the recurrence
(7) T₀(x) = 1, T₁(x) = x, Tₙ₊₁(x) = 2x·Tₙ(x) − Tₙ₋₁(x)
which means that they satisfy
(8) Tₙ(cos θ) = cos nθ
Therefore, if we substitute t = cos α in (5) or x = 2t = 2·cos α in (2) these equations reduce to
(9) cos 6α = cos α
and this trigonometric equation is easily solved. We have
6α = α + 2kπ ⋁ 6α = −α + 2kπ, k ∈ ℤ
5α = 2kπ ⋁ 7α = 2kπ, k ∈ ℤ
α = ²⁄₅·kπ ⋁ α = ²⁄₇·kπ, k ∈ ℤ
so for (2) we have the roots
x₁ = 2·cos 0, x₂ = 2·cos(²⁄₅·π), x₃ = 2·cos(⁴⁄₅·π),
x₄ = 2·cos(²⁄₇·π), x₅ = 2·cos(⁴⁄₇·π), x₆ = 2·cos(⁶⁄₇·π)
Here x₁ = 2·cos 0 = 2 so the left hand side of (2) has a factor x − 2 and x₂ = 2·cos(²⁄₅·π) = −¹⁄₂ + ¹⁄₂√5, x₃ = 2·cos(⁴⁄₅·π) = −¹⁄₂ − ¹⁄₂√5 are the zeros of the quadratic polynomial x² + x − 1 which therefore also is a factor of the left hand side of (2).
The remaining roots x₄ = 2·cos(²⁄₇·π), x₅ = 2·cos(⁴⁄₇·π), x₆ = 2·cos(⁶⁄₇·π) are the roots of the cubic equation
(10) x³ + x² − 2x − 1 = 0
To prove this, we can substitute x = u + 1/u in (10) which results in
(11) u⁶ + u⁵ + u⁴ + u³ + u² + u + 1 = 0
Since the left hand side of (11) is a geometric series with sum (u⁷ − 1)/(u − 1) we can also write this as
(12) (u⁷ − 1)/(u − 1) = 0
Clearly, the roots of (12) and therefore of (11) are the seventh roots of unity except for u = 1 itself, so we have u = exp(k·²⁄₇·π·i), k = 1..6 and since x = u + 1/u this means that the roots of (10) are x = exp(k·²⁄₇·π·i) + exp(−k·²⁄₇·π·i) = 2·cos(k·²⁄₇·π), k = 1..6. Of course, for k = 4..6 we get the same values as for k = 1..3 so we indeed have the three roots x₄ = 2·cos(²⁄₇·π), x₅ = 2·cos(⁴⁄₇·π), x₆ = 2·cos(⁶⁄₇·π) for (10) and the left hand side of (10) is therefore also a factor of the left hand side of (2).
WolframAlpha is incapable of coming up with the solutions x₄ = 2·cos(²⁄₇·π), x₅ = 2·cos(⁴⁄₇·π), x₆ = 2·cos(⁶⁄₇·π) for (10) and instead produces expressions for the solutions of (10) that contain cube roots of complex numbers even though all solutions of (10) are real. This is because the solutions of a cubic equation that has three distinct real but irrational roots cannot be expressed algebraically in radicals without using complex numbers.
This is known as the _casus irreducibilis_ but there is a standard method to solve cubic equations with three distinct real but irrational roots trigonometrically, without using complex numbers. However, the intriguing thing here is that the expressions x₄ = 2·cos(²⁄₇·π), x₅ = 2·cos(⁴⁄₇·π), x₆ = 2·cos(⁶⁄₇·π) for the roots of (10) cannot be obtained either by using the standard trigonometric method to solve a cubic equation with three real roots.
Substituting x = (z − 1)/3 in (10) results in
(13) z³ − 21z − 7 = 0
and solving this depressed cubic trigonometrially in the conventional way we can express the roots of this equation as
z₁ = 2√7·cos(¹⁄₃·arccos(¹⁄₁₄√7))
z₂ = 2√7·cos(¹⁄₃·arccos(¹⁄₁₄√7) − ²⁄₃·π)
z₃ = 2√7·cos(¹⁄₃·arccos(¹⁄₁₄√7) + ²⁄₃·π)
So, since x = (z − 1)/3, we can express the roots of (10) as
x₄ = −¹⁄₃ + ²⁄₃√7·cos(¹⁄₃·arccos(¹⁄₁₄√7))
x₅ = −¹⁄₃ + ²⁄₃√7·cos(¹⁄₃·arccos(¹⁄₁₄√7) − ²⁄₃·π)
x₆ = −¹⁄₃ + ²⁄₃√7·cos(¹⁄₃·arccos(¹⁄₁₄√7) + ²⁄₃·π)
but it is not evident at all how these expressions can be reduced to the much simpler expressions
x₄ = 2·cos(²⁄₇·π)
x₅ = 2·cos(⁴⁄₇·π)
x₆ = 2·cos(⁶⁄₇·π)
Since we have proved that x − 2, x² + x − 1 and x³ + x² − 2x − 1 are all factors of the left hand side of (2) and since there are no other factors since any root of (2) is a zero of one of these factors it follows that we can write (2) as
(14) (x − 2)(x² + x − 1)(x³ + x² − 2x − 1) = 0
Since the set of roots of (2) consist of the union of the set of roots of (1) and the set of roots of (3) we still need to find out which roots of (2) are roots of (1) and which roots of (2) are roots of (3). To do this, we can apply the double angle identity for the cosine to (9) which gives
(15) 2·cos² 3α − 1 = 2·cos² ½α − 1
so we have cos² 3α = cos² ½α and therefore (9) is equivalent with
(16) cos 3α = cos ½α ⋁ cos 3α = −cos ½α
where the first equation is also obtained by substituting x = 2·cos α in (1) and the second by substituting this in (3) subject to the condition 0 ≤ α ≤ π to ensure the proper sign of √(x + 2) = cos ½α.
Solving cos 3α = cos ½α gives α = ⁴⁄₅·kπ ⋁ α = ⁴⁄₇·kπ, k ∈ ℤ so taking the condition 0 ≤ α ≤ π into account we have the roots x₁ = 2·cos 0 = 2, x₃ = 2·cos(⁴⁄₅·π) = −¹⁄₂ − ¹⁄₂√5, x₅ = 2·cos(⁴⁄₇·π) for (1).
Similarly, solving cos 3α = −cos ½α gives α = ²⁄₇·(2k + 1)π ⋁ α = ²⁄₅·(2k − 1)π, k ∈ ℤ and taking the condition 0 ≤ α ≤ π into account this gives the roots x₂ = 2·cos(²⁄₅·π) = −¹⁄₂ + ¹⁄₂√5, x₄ = 2·cos(²⁄₇·π), x₆ = 2·cos(⁶⁄₇·π) for (3).
Thanks a lot my friend for sharing this solution👍👍👍
wtf?
What
TF?