The solutions to the polynomial equation ( n^2 + n^4 = n^6 ) include ( n = 0 ), approximately ( n = 1.272 ), approximately ( n = -1.272 ), and the complex solutions approximately ( n = 0.786 i ) and ( n = -0.786 i ). However, only ( n = 1.272 ) (the square root of the golden ratio, ( phi ) is a viable solution in the context of the Pythagorean theorem, as negative solutions, complex solutions, and ( n = 0 ) are not optimal for representing the lengths of the sides of a triangle. I hope this cleared up any confusion.
Fun fact- The Golden Ratio is the most irrational n.o, hence places where irrationality is expected and will work the best, use golden ratio automatically(not intentionally); not because of magic.
@@opalkarli Its rational approximations converge slowest in terms of the magnitude of the numerator and denominator. Has to do with the fact that its continued fraction representation equals 1;1,1,1,1,...
Let n = 1.616, then a² = 2.618 and a³ = 4.236. The hypotenuse of the rectangle triangle with perpendicular sides a² and a³ es just sqrt(n² + n² n²) wich gives 3.07768. The equation n² + n⁴ = n⁶ is jus a polynomial in n with two solutions: 0 and 1.618 (the golden ratio) The way you put the problem is wrong for there is not such a triangle.
It's a fairly straightforward problem for a "Math Olympiad, or even a Harvard University Entrances Exam. It was clear from the start it was going to be a quadratic multiplies by n^2. I was even able to work out n^2 in my head using the quadratic equation.
By creating problem where the answer is the square root of the golden ratio, we have created a problem that has one solution of the square root of the golden ratio. In other words, we forced the answer by choosing the side lengths. Isn't that so impressive? The answer is of course not.
@@tonioliii точка не считается отрезком. При x→0 отрезок ещё может быть бесконечно малым, но фигура с нулевым габаритом - это точка по определению, т. е. уже не отрезок. No exceptions.☝👀
as there cannot be real certain 95 angles degrees, there sno exact solution of this problem, so ı wont be getting tired by trying to think about thşs math solution which gets me too tired! ( ı hate thinking! it is so disturbing)
The solutions to the polynomial equation ( n^2 + n^4 = n^6 ) include ( n = 0 ), approximately ( n = 1.272 ), approximately ( n = -1.272 ), and the complex solutions approximately ( n = 0.786 i ) and ( n = -0.786 i ). However, only ( n = 1.272 ) (the square root of the golden ratio, ( phi ) is a viable solution in the context of the Pythagorean theorem, as negative solutions, complex solutions, and ( n = 0 ) are not optimal for representing the lengths of the sides of a triangle. I hope this cleared up any confusion.
Fun fact- The Golden Ratio is the most irrational n.o, hence places where irrationality is expected and will work the best, use golden ratio automatically(not intentionally); not because of magic.
how can a number be more irrational than another number
@@opalkarli Its rational approximations converge slowest in terms of the magnitude of the numerator and denominator.
Has to do with the fact that its continued fraction representation equals 1;1,1,1,1,...
@ oh, ok! Thanks!
I love the solution. For some ineffable reason, it doesn't totally surprise me.
n= sqrt(golden ratio)
It only takes one second to realize that the golden ratio is involved.
n^2+(n^2)^2=(n^3)^2
n^2+(n^2)^2=(n^2)^3
1+n^2=(n^2)^2
n^2=Φ
How much do I despise the AI voice over??.no one actually talks like this.
Slightly different method.
n^2 + (n^2)^2 = (n^3)^2
n^2 + (n^2)^2 = n^(3*2)
n^2 + (n^2)^2 = n^(2*3)
n^2 + (n^2)^2 = (n^2)^3
Let u = n^2
u + u^2 = u^3
(u + u^2) - (u + u^2) = u^3 - (u + u^2)
0 = u^3 - u - u^2
0 = u^3 - u^2 - u
u^3 - u^2 - u = 0
u(u^2 - u - 1) = 0
Suppose u = 0
u = 0
Remember, u = n^2
n^2 = 0
sqrt(n^2) = +/- sqrt(0)
n = +/- 0
n = 0 (Yes, a triangle can have such dimensions)
Suppose 1*u^2 - 1*u - 1 = 0
Let a = 1, b = -1, c = -1
u = (-b +/- sqrt[b^2 - 4*a*c]) / (2*a)
u = (-[-1] +/- sqrt[(-1)^2 - 4*1*(-1)]) / (2*1)
u = (1 +/- sqrt[1 + 4]) / (2)
u = (1 +/- sqrt[5]) / 2
u = (1 + sqrt[5]) / 2, or u = (1 - sqrt[5]) / 2
Remember, u = n^2
n^2 = (1 + sqrt[5]) / 2, or n^2 = (1 - sqrt[5]) / 2
(1 - sqrt[5]) / 2 < 0, so reject (1 - sqrt[5]) / 2
n^2 = (1 + sqrt[5]) / 2
sqrt(n^2) = +/- sqrt([1 + sqrt(5)] / 2)
n = +/- sqrt([1 + sqrt(5)] / 2)
n = sqrt([1 + sqrt(5)] / 2), or n = -sqrt([1 + sqrt(5)] / 2)
n = -sqrt([1 + sqrt(5)] / 2) < 0, so reject -sqrt([1 + sqrt(5)] / 2)
n = sqrt([1 + sqrt(5)] / 2)
n = sqrt(1 + sqrt[5]) / sqrt(2)
n1 = 0
n2 = sqrt(1 + sqrt[5]) / sqrt(2)
underrated channel
Let n = 1.616, then a² = 2.618 and a³ = 4.236. The hypotenuse of the rectangle triangle with perpendicular sides a² and a³ es just sqrt(n² + n² n²) wich gives 3.07768. The equation n² + n⁴ = n⁶ is jus a polynomial in n with two solutions: 0 and 1.618 (the golden ratio) The way you put the problem is wrong for there is not such a triangle.
Can’t we solve using trigonometry? Write sin and cos functions of an angle and then equating sum of squares of sin and cos to one?
Можно было бы сразу рассмотреть подобный треугольник с коэффициентом подобия n и сторонами 1, n, n^2.
It's a fairly straightforward problem for a "Math Olympiad, or even a Harvard University Entrances Exam. It was clear from the start it was going to be a quadratic multiplies by n^2. I was even able to work out n^2 in my head using the quadratic equation.
Thanks for the challenge. I solved in less than a minute by using the quadratic formula.
The triangle can be simplified by dividing all sides by n. the resulting triangle is solved by pythagoras theorem easily.
👍 right
That's what x=n^2 does
X,2×+5=8[n3]
By creating problem where the answer is the square root of the golden ratio, we have created a problem that has one solution of the square root of the golden ratio. In other words, we forced the answer by choosing the side lengths. Isn't that so impressive? The answer is of course not.
Any reliable source that phi systematically appears in the spirals of galaxies? (Apart from coincidental matches.)
Isnt N² and N The same length?
equation should have been n^2(n^3-N^2-1)=0
No. n^6 / n^2 = n^(6-2) = n^4.
9×9-4×4=65squroth=8.0622
Sooo Goood in the end was so melodic 😂😂❤
😂
n²:=x. x+x²=x³. x+x²-x³=x(1+x-x²)=0. x=0 - точка, не считается. 1+x-x²=0⇔x²-x-1=0. D=5, √D=√5. x=(1±√5)/2. n=√[(1±√5)/2]. (1-√5)
Thanks Casey.
Golden ratio yet again. I'm not sure whether I should be excited or disappointed.
Why disappointed?
X,2×+5=5=8
I am not shocked! Try harder next time 😊
I don't see why you can't have a triangle with height 0 and base 0^2.
🥲✌️
🪙Golden Ratio= 1.618 💛
That was simply amazing ❤❤. Yesterday only I was reading about golden ratio and this question popped up today! Coincidence?
It's destiny! 🤩
N=0
Not counts. )
@zawatsky yes counts
@@tonioliii точка не считается отрезком. При x→0 отрезок ещё может быть бесконечно малым, но фигура с нулевым габаритом - это точка по определению, т. е. уже не отрезок. No exceptions.☝👀
@@tonioliiiit would not be a triangle if n = 0 or less
@@zawatsky Mathematically it counts, in the case of degenerate triangles/
N^2+N^4=(N^3)^2
Sigma sigma boy sihma boy
Not shocked
Haha, cat...
The cat approves 😘
😮
Да
1
φ
as there cannot be real certain 95 angles degrees, there sno exact solution of this problem, so ı wont be getting tired by trying to think about thşs math solution which gets me too tired! ( ı hate thinking! it is so disturbing)
1