Solving A Quartic #algebra

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[GillesF]

I did this way (step by step, a long way) ...

x⁵ = (x - 1)⁵

(x - 1)⁵ = x⁵

((x - 1)⁵)/x⁵ = x⁵/ x⁵

((x - 1)/x)⁵ = 1

set k = (x - 1)/x

k⁵ = 1

---

/// "roots of unity" of k⁵ = 1 (recall):

// note:

• argument: n·(π/5) (n = [2..10] by 2)

• module: 1 (for each)

• polar form: 1·e^(i·n·(π/5)) = e^(i·n·(π/5))

// computation:

• root #1 (of unity): e^(i·(2π/5)) = cos(2π/5) + i·sin(2π/5) ≈ 0.309 + 0.951·i

• root #2 (of unity): e^(i·(4π/5)) = cos(4π/5) + i·sin(4π/5) ≈ -0.809 + 0.587·i

• root #3 (of unity): e^(i·(6π/5)) = cos(6π/5) + i·sin(6π/5) ≈ -0.809 - 0.587·i

• root #4 (of unity): e^(i·(8π/5)) = cos(8π/5) + i·sin(8π/5) ≈ 0.309 - 0.951·i

• root #5 (of unity): e^(i·2π) = 1

---

/// case: (x - 1)/x = 0.309 + 0.951·i

(x - 1)/x = 0.309 + 0.951·i

x/x - 1/x = 0.309 + 0.951·i

1 - 1/x = 0.309 + 0.951·i

1/x = 1 - 0.309 + 0.951·i

recall: 1/a = b => a = 1/b

x = 1/(1 - 0.309 + 0.951·i)

x = 1/(0.691 + 0.951·i)

recall:

||
|| division between 2 complex numbers:
||
|| a + bi (a + bi)·(c - di) (ac + bd) + i·(bc - ad)
|| = =
|| c + di (c + di)·(c - di) c² + d²
||

x = (1 + 0·i)/(0.691 + 0.951·i)

note:

• a = 1

• b = 0

• c = 0.691

• d = 0.951

x = [(ac + bd) + i·(bc - ad)]/(c² + d²)

x = [(1·0.691 + 0·0.951) + i·(0·0.691 - 1·0.951)]/(0.691² + 0.951²)

■ x ≈ 0.5 - 0.688·i

---

/// case: (x - 1)/x = -0.809 + 0.587·i

(x - 1)/x = -0.809 + 0.587·i

x/x - 1/x = -0.809 + 0.587·i

1 - 1/x = -0.809 + 0.587·i

1/x = 1 + 0.809 + 0.587·i

recall: 1/a = b => a = 1/b

x = 1/(1 + 0.809 + 0.587·i)

x = 1/(1.809 + 0.587·i)

recall:

||
|| division between 2 complex numbers:
||
|| a + bi (a + bi)·(c - di) (ac + bd) + i·(bc - ad)
|| = =
|| c + di (c + di)·(c - di) c² + d²
||

x = (1 + 0·i)/(1.809 + 0.587·i)

note:

• a = 1

• b = 0

• c = 1.809

• d = 0.587

x = [(ac + bd) + i·(bc - ad)]/(c² + d²)

x = [(1·1.809 + 0·0.587) + i·(0·1.809 - 1·0.587)]/(1.809² + 0.587²)

■ x ≈ 0.5 - 0.162·i

---

/// case: (x - 1)/x = -0.809 - 0.587·i

SAME REASONING AS PREVIOUSLY

x = [(1·1.809 - 0·0.587) + i·(0·1.809 + 1·0.587)]/(1.809² + 0.587²)

■ x ≈ 0.5 + 0.162·i


---

/// case: (x - 1)/x = 0.309 - 0.951·i

SAME REASONING AS PREVIOUSLY

x = [(1·0.691 + 0·0.951) + i·(0·0.691 + 1·0.951)]/(0.691² + 0.951²)

■ x ≈ 0.5 + 0.688·i

---

/// case: (x - 1)/x = 1

(x - 1)/x = 1

x/x - 1/x = 1

1 - 1/x = 1

1/x = 1 - 1

1/x = 0

NO SOLUTION !

---

/// final results:

■ root #1: x ≈ 0.5 - 0.688·i

■ root #2: x ≈ 0.5 - 0.162·i

■ root #3: x ≈ 0.5 + 0.162·i

■ root #4: x ≈ 0.5 + 0.688·i

🙂

[black_eagle]

x^5 = (x-1)^5 => ((x-1)/x)^5 = 1 or u^5 =1 where u = (x-1)/x

u^5 = e^(i*2pi*n) => u = e^(i*2pi*n/5), n = 0, 1, 2, 3, 4

u=(x-1)/x = 1 - 1/x => x = 1/(1-u) = 1/(1 - e^(i*2pi*n/5)), n = 1, 2, 3, 4.

Expanding the denominator and putting in standard form gives x = 1/2 + i * 1/2 * sin(2pi*n/5) / (1 - cos(2pi*n/5))

You can substitute the exact values of these sines and cosines and simplify but I just left it like this.

[shannonmcdonald]

Please remind me: when we move a curve in the positive direction along the x axis, why do we subtract the number of places instead of adding them?

[ignaciosavi]

No Solutions in real numbers my first guess

[danielpraise]

x⁵ = (x-1)⁵
I would like to let
(x-1) = y
x - y = 1
x⁵ - y⁵ = 0
x² + y² = 1 + 2xy
x³ - y³ = (1 + 3xy)

(x³-y³)(x²+y²) = (1+2xy)(1+3xy)

x⁵ + x³y² - y³x² - y⁵ = 1 + 3xy + 2xy + 6(xy)²
x⁵ - y⁵ + x²y²(x-y) = 1+5xy + (6xy)²
Let xy = a
5a² + 5a + 1 = 0
a = -5 ± √(25 - 20)/10
a = -5 ± √5/10
x - y = 1
xy = (-5 ± √5)/10
Solving the system u get the value of x