(x + y) (x² + y²) + 2 = 0 has still to be solved. Should deliver 3 solutions with 3 * 4 = 12 solutions overall! Our 4 plus 12 = 16. ✅ We can also write it as (x + y) (x² + y²) = -2. So we need to find two numbers, so that the product is -2. If we find a pair, the rest should be easy. But let's simplify first: x² + y² = (x + iy) (x - iy) = x² - (iy)² = x² - (-y²). ✅ This didn't help much, but it's good to see the geometry. Let's assume, x + y = z = c + 0i real. So x² + y² = -2 / c also real. z² = x² + 2xy + y² also real. => xy also real! y = s * conj(x). But x² + y² was real, so s = 0 (and a = 0 ∨ b = 0) or s = 1. Case s = 0: y = 0, x^3 = -2. => x = - ∛2 ✅(∨ x = ∛2 (½ + ½√3 i) ✅∨ x = ∛2 (½ - ½√3 i) ✅). Symmetry: x and y can be switched without changing the result! F.e. x = 0 and y = - ∛2 is also a valid solution. ✅ Case s = 1: y = conj(x) = |x|² / x. x = a + b i, y = a - b i, 2xy = 2 |x|² = 2 (a² + b²). iy = b + a i, x + y = 2a = c. -2 / c = -1 / a = x² + y² = 2 a² - 2 b² = 2 (a + b) (a - b). b = 0, 2 a³ = -1 => x = y = - 1/∛2 solves this. ✅ All the x = y solutions lead back to x^4 - 1 = 0 = y^4 - 1 system from the video.
I think we can observe that X=1 is a solution and X=-1 is also a solution. This means we can divide by x^2-1. This justifies splitting 2 into 1+1 and moving one of the 1s to the left hand side (X^4+x-1)^4-1=1-x^4 ((X^4+x-1)^2-1)((X^2+x-1)^2+1)=-(x^2-1)(x^2+1) (X^4+X)(X^4+x-2)((X^4+x-1)^2+1)+(X-1)(X+1)(X^2+1)=0 X(X+1)(x^2-x+1)(x-1)(x^3+x^2+X+1+1)(...)+(X+1)(x-1)(x^2+1)=0 we know X=1 and X=-1 are solutions so we can look for the other solutions from the remainder polynomial X(x^2-x+1)(x^3+x^2+X+2)((x^4+x-1)^2+1)+x^2+1=0 X^2-x+1=(x-1/2)^2+3)4>0 and the last paranthesis is always >0 So the only solution is possible when X*(x^3+x^2+X+2)
problem (x⁴ + x -1)⁴ = 2 - x⁴ According to the ingenious solution method of substitution in the video, (x⁴ + x -1)⁴ + x⁴ = 2 (x⁴ + x -1)⁴ + x⁴ + x - 1 = 2 + x - 1 = x + 1 x = (x⁴ + x -1)⁴ + (x⁴ + x - 1) - 1 Make the substitution y = x⁴ + x -1 So we now have the system y = x⁴ + x -1 x = y⁴ + y - 1 Subtract to get y-x = x⁴-y⁴ + x -y x⁴-y⁴ + 2(x -y) = 0 (x²-y²)(x² +y²) + 2(x-y) = 0 (x-y)(x+y)(x² +y²)+2(x-y) = 0 (x-y) [ (x+y)(x² +y²)+2 ] = 0 By the zero product property x-y = 0 x = y x = x⁴ + x - 1 x⁴-1 = 0 (x²-1)(x² +1) = 0 (x-1)(x+1)(x-i)(x+i) = 0 Roots from this equation are x = -1, 1, -i, i To add to what was in the video, the remaining equation using that system is (x+y)(x²+y²)+2 = 0 (x+y)(x² +y²) = -2 It has been verified that setting each factor to a factor of -2 does not solve this system. I tried several factors of -2 to no avail. answer x ∈ { -1, 1, -i, i, }
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There are 12 missing solutions, too bad!
I actually guessed successfully the 4 solutions +-1 and +-i before I even solved. Crazy 🤣
You are so bubbly and enthusiastic, I love it! I too adore maths.
Very cool! We could feel your growing enthusiasm! Thanks!
Glad you liked it! 🤩
what about the second factor?
That’s for you to figure out 🤪
He has no idea!
Cool!
Hey, is it okay if you do my request next? I put it in the comment section of the last video.
Very nice
Thank you!
y = f(x), x = f(y)... wait! f(f(x)) = id(x) = x ? Nice functional equation... and then back to f(x) = x^4 + x - 1...
Sir could you please tell me a math keyboard app from which I can write math symbols like 3^2n on youtube thumbnail?
You can use this: lingojam.com/SuperscriptGenerator
Why don't you subtract x^4 on boath sides?
Does that help?
@@SyberMath Yes. You can factorise left side and get one of the factors (x^4 - 1) and on the right you have the same thing exposing -2.
x = 1 or -1
I could easily say x =- 1 and x=1 are the answers by substitutions.
Good job!
(x + y) (x² + y²) + 2 = 0 has still to be solved. Should deliver 3 solutions with 3 * 4 = 12 solutions overall! Our 4 plus 12 = 16. ✅
We can also write it as (x + y) (x² + y²) = -2. So we need to find two numbers, so that the product is -2. If we find a pair, the rest should be easy. But let's simplify first: x² + y² = (x + iy) (x - iy) = x² - (iy)² = x² - (-y²). ✅ This didn't help much, but it's good to see the geometry.
Let's assume, x + y = z = c + 0i real. So x² + y² = -2 / c also real. z² = x² + 2xy + y² also real. => xy also real! y = s * conj(x).
But x² + y² was real, so s = 0 (and a = 0 ∨ b = 0) or s = 1.
Case s = 0: y = 0, x^3 = -2. => x = - ∛2 ✅(∨ x = ∛2 (½ + ½√3 i) ✅∨ x = ∛2 (½ - ½√3 i) ✅).
Symmetry: x and y can be switched without changing the result! F.e. x = 0 and y = - ∛2 is also a valid solution. ✅
Case s = 1: y = conj(x) = |x|² / x. x = a + b i, y = a - b i, 2xy = 2 |x|² = 2 (a² + b²). iy = b + a i, x + y = 2a = c.
-2 / c = -1 / a = x² + y² = 2 a² - 2 b² = 2 (a + b) (a - b). b = 0, 2 a³ = -1 => x = y = - 1/∛2 solves this. ✅
All the x = y solutions lead back to x^4 - 1 = 0 = y^4 - 1 system from the video.
Все красиво, но... неужели нельзя не трещать как пулемёт, а говорить с толком и расстоновкой?
Чел, я его смотрю на двойной скорости, и все норм 👍🏻 Так что дело в тебе.
I think we can observe that X=1 is a solution and X=-1 is also a solution. This means we can divide by x^2-1. This justifies splitting 2 into 1+1 and moving one of the 1s to the left hand side
(X^4+x-1)^4-1=1-x^4
((X^4+x-1)^2-1)((X^2+x-1)^2+1)=-(x^2-1)(x^2+1)
(X^4+X)(X^4+x-2)((X^4+x-1)^2+1)+(X-1)(X+1)(X^2+1)=0
X(X+1)(x^2-x+1)(x-1)(x^3+x^2+X+1+1)(...)+(X+1)(x-1)(x^2+1)=0 we know X=1 and X=-1 are solutions so we can look for the other solutions from the remainder polynomial
X(x^2-x+1)(x^3+x^2+X+2)((x^4+x-1)^2+1)+x^2+1=0
X^2-x+1=(x-1/2)^2+3)4>0 and the last paranthesis is always >0
So the only solution is possible when
X*(x^3+x^2+X+2)
Do **NOT** use X and x together! They represent *different* variables. Stick exclusively with either the lower case or the upper case.
@@robertveith6383Lol
Secant method (calculated online)
-1,189012...
-0,19032...
problem
(x⁴ + x -1)⁴ = 2 - x⁴
According to the ingenious solution method of substitution in the video,
(x⁴ + x -1)⁴ + x⁴ = 2
(x⁴ + x -1)⁴ + x⁴ + x - 1 = 2 + x - 1
= x + 1
x = (x⁴ + x -1)⁴ + (x⁴ + x - 1) - 1
Make the substitution
y = x⁴ + x -1
So we now have the system
y = x⁴ + x -1
x = y⁴ + y - 1
Subtract to get
y-x = x⁴-y⁴ + x -y
x⁴-y⁴ + 2(x -y) = 0
(x²-y²)(x² +y²) + 2(x-y) = 0
(x-y)(x+y)(x² +y²)+2(x-y) = 0
(x-y) [ (x+y)(x² +y²)+2 ] = 0
By the zero product property
x-y = 0
x = y
x = x⁴ + x - 1
x⁴-1 = 0
(x²-1)(x² +1) = 0
(x-1)(x+1)(x-i)(x+i) = 0
Roots from this equation are
x = -1, 1, -i, i
To add to what was in the video, the remaining equation using that system is
(x+y)(x²+y²)+2 = 0
(x+y)(x² +y²) = -2
It has been verified that setting each factor to a factor of -2 does not solve this system.
I tried several factors of -2 to no avail.
answer
x ∈ {
-1,
1,
-i,
i,
}