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Nice math video, you the best sir.
Thanks and welcome. Keep watching
6 + 1/x = u => x = 1/(u - 6)(u + 1)⁴ - (u - 1)⁴ = 2402(4u³ + 4u) = 240u³ + u - 30 = 0(u³ - 27) + (u - 3) = 0(u - 3)(u² + 3u + 10) = 0u = 3 => *x = -1/3*u² + 3u + 10 = 0 => complex roots
verification (7 - 3)⁴ - (5 - 3)⁴ = 2404⁴ - 2⁴ = 2402⁸ - 2⁴ = 240256 - 16 = 240240 = 240
Awesome 😎
Ingenioso
Thanks 👍💯😊 for your input and support
Good, but why you didn't find complex roots?
I assumed that they were NOT required
[7 + (1/x)]⁴ - [5 + (1/x)]⁴ = 240 → let: a = 1/x(7 + a)⁴ - (5 + a)⁴ = 240[(7 + a)²]² - [(5 + a)²]² = 240[(7 + a)² + [(5 + a)²].[(7 + a)² - (5 + a)²] = 240[(49 + 14a + a²) + (25 + 10a + a²)].[(49 + 14a + a²) - (25 + 10a + a²)] = 240[49 + 14a + a² + 25 + 10a + a²].[49 + 14a + a² - 25 - 10a - a²] = 240[74 + 24a + 2a²].[24 + 4a] = 2402.[37 + 12a + a²].4.[6 + a] = 240(37 + 12a + a²).(6 + a) = 30222 + 37a + 72a + 12a² + 6a² + a³ = 30a³ + 18a² + 109a + 192 = 0a³ + 15a² + 3a² + 45a + 64a + 192 = 0a³ + 15a² + 64a + 3a² + 45a + 192 = 0(a³ + 15a² + 64a) + (3a² + 45a + 192)= 0a.(a² + 15a + 64) + 3.(a² + 15a + 64) = 0(a + 3).(a² + 15a + 64) = 0First case: (a + 3) = 0a + 3 = 0→ a = - 3Second case: (a² + 15a + 64) = 0a² + 15a + 64 = 0Δ = (15)² - (4 * 64) = 225 - 256 = - 31 → no real solutionResume: a = - 3Recall: a = 1/x → x = 1/a→ x - 1/3If you want to continu with complex numbers Second case: (a² + 15a + 64) = 0a² + 15a + 64 = 0Δ = (15)² - (4 * 64) = 225 - 256 = - 31 = 31i²a = (- 15 ± i√31)/2Recall: a = 1/x → x = 1/ax = 2/(- 15 ± i√31)First value: x = 2/(- 15 + i√31)x = 2.(- 15 - i√31)/[(- 15 + i√31).(- 15 - i√31)]x = 2.(- 15 - i√31)/[225 + 31]x = 2.(- 15 - i√31)/256→ x = - (15 + i√31)/128Second value: x = 2/(- 15 - i√31)x = 2.(- 15 + i√31)/[(- 15 - i√31).(- 15 + i√31)]x = 2.(- 15 + i√31)/[225 + 31]x = 2.(- 15 + i√31)/256→ x = - (15 - i√31)/128
cheating videos😢
Nowhere in problem is it said that x has only real value ? So, why is the value of x not imaginary?😊
I restricted the solution to real values
Nice math video, you the best sir.
Thanks and welcome. Keep watching
6 + 1/x = u => x = 1/(u - 6)
(u + 1)⁴ - (u - 1)⁴ = 240
2(4u³ + 4u) = 240
u³ + u - 30 = 0
(u³ - 27) + (u - 3) = 0
(u - 3)(u² + 3u + 10) = 0
u = 3 => *x = -1/3*
u² + 3u + 10 = 0 => complex roots
verification
(7 - 3)⁴ - (5 - 3)⁴ = 240
4⁴ - 2⁴ = 240
2⁸ - 2⁴ = 240
256 - 16 = 240
240 = 240
Awesome 😎
Ingenioso
Thanks 👍💯😊 for your input and support
Good, but why you didn't find complex roots?
I assumed that they were NOT required
[7 + (1/x)]⁴ - [5 + (1/x)]⁴ = 240 → let: a = 1/x
(7 + a)⁴ - (5 + a)⁴ = 240
[(7 + a)²]² - [(5 + a)²]² = 240
[(7 + a)² + [(5 + a)²].[(7 + a)² - (5 + a)²] = 240
[(49 + 14a + a²) + (25 + 10a + a²)].[(49 + 14a + a²) - (25 + 10a + a²)] = 240
[49 + 14a + a² + 25 + 10a + a²].[49 + 14a + a² - 25 - 10a - a²] = 240
[74 + 24a + 2a²].[24 + 4a] = 240
2.[37 + 12a + a²].4.[6 + a] = 240
(37 + 12a + a²).(6 + a) = 30
222 + 37a + 72a + 12a² + 6a² + a³ = 30
a³ + 18a² + 109a + 192 = 0
a³ + 15a² + 3a² + 45a + 64a + 192 = 0
a³ + 15a² + 64a + 3a² + 45a + 192 = 0
(a³ + 15a² + 64a) + (3a² + 45a + 192)= 0
a.(a² + 15a + 64) + 3.(a² + 15a + 64) = 0
(a + 3).(a² + 15a + 64) = 0
First case: (a + 3) = 0
a + 3 = 0
→ a = - 3
Second case: (a² + 15a + 64) = 0
a² + 15a + 64 = 0
Δ = (15)² - (4 * 64) = 225 - 256 = - 31 → no real solution
Resume: a = - 3
Recall: a = 1/x → x = 1/a
→ x - 1/3
If you want to continu with complex numbers
Second case: (a² + 15a + 64) = 0
a² + 15a + 64 = 0
Δ = (15)² - (4 * 64) = 225 - 256 = - 31 = 31i²
a = (- 15 ± i√31)/2
Recall: a = 1/x → x = 1/a
x = 2/(- 15 ± i√31)
First value: x = 2/(- 15 + i√31)
x = 2.(- 15 - i√31)/[(- 15 + i√31).(- 15 - i√31)]
x = 2.(- 15 - i√31)/[225 + 31]
x = 2.(- 15 - i√31)/256
→ x = - (15 + i√31)/128
Second value: x = 2/(- 15 - i√31)
x = 2.(- 15 + i√31)/[(- 15 - i√31).(- 15 + i√31)]
x = 2.(- 15 + i√31)/[225 + 31]
x = 2.(- 15 + i√31)/256
→ x = - (15 - i√31)/128
cheating videos😢
Nowhere in problem is it said that x has only real value ? So, why is the value of x not imaginary?😊
I restricted the solution to real values