Solving A Nice Polynomial Equation

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PLAYLISTS 🎵 :

[jaimeduncan]

I did it differently, same results I used the difference of squares X^4 - (x-1)^4=0 => (x^2 -(x-1)^2)(x^2 +(x-1)^2) =0 then set each term equal to zero tthe first one will be a linea equation (you can use difference again or expand) and the second one a quadratic, solve them and get the same values.

[brianvincent]

You already did this one but I'll watch anyway 😅

[xwyl]

Draw the thing on the complex plane and immediately see that x-1 and x have the same length, same imaginary part and therefore opposite real parts,
To satisfy the equation, the angle difference between x-1 and x must be multiple of pi/2
Then obviously there are 3 solutions: 1/2, 1/2+i/2, 1/2-i/2. The equation is of the 3rd power, so these 3 solutions are all there are.

[vitalsbat]

Test x=0, it does not work
Thus dividing both sides by x⁴ is fine.
(1-1/x)⁴ = 1
1-1/x = 1 or -1 or i or -i
Then its linear, giving three solutions (1 gives undefined)
x = 1/2, x = 1/2 + i/2 and x = 1/2 - i/2

[onegreengoat]

Very nice. This could almost go in your a+bi channel because I thought about the symmetry of the roots of unity. e^i(pi/4) raised to the 4th power = e^i(3pi/4) raised to the 4th power = e^i(-3pi/4) raised to the 4th power = e^i(-pi/4) raised to the 4th power. Scaling these numbers by 1/sqrt(2) you have (1/sqrt(2))e^i(pi/4) - 1 = (1/sqrt(2))e^i(3pi/4) and (1/sqrt(2))e^i(-pi/4) - 1 = (1/sqrt(2))e^i(-3pi/4). I like this problem

[MrGeorge]

x⁴ - (x - 1)⁴ = 0 is a difference of two squares:

a⁴ - b⁴ = (a² + b²) (a² - b²) = (a² + b²) (a + b) (a - b)

so with a = x and b = x - 1

x⁴ - (x - 1)⁴ = 0 = (x² + (x - 1)²) (x + (x - 1)) (x - (x - 1)) = (2x² - 2x + 1) (2x - 1) (1)

[Nobodyman]

I think if you do this: x⁴-(x-1)⁴=0, this eq is solves faster

[DonEnsley]

problem

(x-1)⁴ = x⁴

Subtract x⁴ from both sides.

(x-1)⁴ - x⁴ = 0

Expanded, this is

-4x³ + 6x² -4x + 1 = 0

which is a cubic, so we expect 3 roots.

It is also a difference of 2 squares. Factor.

(x-1)⁴ - x⁴ = 0
[(x-1)² - x²][(x-1)² + x²] = 0
(x-1-x)(x-1+x)[(x-1)² + x²] = 0
-1 (2x - 1)[(x-1)² + x²] = 0
(2x - 1)[(x-1)² + x²] = 0

By the 0 product principle,

(2x - 1) = 0
x = ½

(x-1)² + x² = 0
2x²- 2x + 1 = 0

By the quadratic formula,

x = ½ (1+i), ½ (1-i)

answer

x ϵ { ½, ½ (1-i), ½ (1+i) }

[kassuskassus]

The second method is nicer and faster.

[jimschneider]

I am afraid I've already seen this one on another channel about two weeks ago, except that one was x^6 = (x - 1)^6 (which also has only one real solution). But now I'm curious - is there a positive integer n such that x^(2*n) = (x - 1)^(2*n) has a real solution other than x = 1/2?

[migry]

I can't imagine what the solution might look like.

[mynameisgood]

Can you just use that a^2 - b^2 = (a-b)(a+b)?