Obviously if you expand the right side you wind up with an a^4 that will cancel the one on the left. So this will just be a cubic polynomial, and that's readily solvable. On top of that, it's immediately clear that a = 0.5 is a real solution, so we could divide that out of the cubic and have a quadratic left. But, there is a better way to proceed. Do this: a^4 = (a-1)^4 1 = (a-1)^4 / a^4 1 = ((a-1)/a)^4 1 = (1-1/a)^4 So now we see that 1-1/a is the fourth roots of unity. So: 1 - 1/a = 1 1 - 1/a = i 1 - 1/a = -1 1 - 1/a = -i The first one of these leads to infinity for a - that's the one that goes away to leave us with just three solutions, as we decided right off the bat we needed (because the a^4 cancels from both sides). Now we can just solve the other three: a = 1/(1-i) a = 1/2 a = 1/(1+i) You can put those into a more normalized form if you want, with the i's in the numerators, but those are the solutions.
■ a = 1/2 + i/2 ■ a = 1/2 - i/2 => I T ' S A S Y O U R S O L U T I O N B E C A U S E I T ' S A V E R Y N I C E W A Y T O S O L V E T H E E Q U A T I O N . 🙂🙂🙂
And at the end of the day who really cares. And if it takes her that length of time to solve a fairly straightforward equation have all the students gone to sleep😂
Ja, schon klar. Wurzel aus einer negativen Zahl.... Bei diesem i-Blödsinn steigt doch jeder normal denkende Mensch aus. Schon die Aufgabe selbst a = a - 1 zeigt doch, daß es hier keine vernünftige Lösung geben kann.
Obviously if you expand the right side you wind up with an a^4 that will cancel the one on the left. So this will just be a cubic polynomial, and that's readily solvable. On top of that, it's immediately clear that a = 0.5 is a real solution, so we could divide that out of the cubic and have a quadratic left.
But, there is a better way to proceed. Do this:
a^4 = (a-1)^4
1 = (a-1)^4 / a^4
1 = ((a-1)/a)^4
1 = (1-1/a)^4
So now we see that 1-1/a is the fourth roots of unity. So:
1 - 1/a = 1
1 - 1/a = i
1 - 1/a = -1
1 - 1/a = -i
The first one of these leads to infinity for a - that's the one that goes away to leave us with just three solutions, as we decided right off the bat we needed (because the a^4 cancels from both sides). Now we can just solve the other three:
a = 1/(1-i)
a = 1/2
a = 1/(1+i)
You can put those into a more normalized form if you want, with the i's in the numerators, but those are the solutions.
Thank you so much for sharing this
Thank you so much sir
a⁴ = (a − 1)⁴
a⁴ = a⁴ − 4a³ + 6a² − 4a + 1
0 = −4a³ + 6a² − 4a + 1
a³ − 3a²/2 + a − 1/4 = 0
(a − 1/2) ⋅ (a² − a + 1/2) = 0
a₁ = 1/2
a₂,₃ = 1/2 ± √(1/4 − 1/2)
= 1/2 ± √(−1/4)
= 1/2 ± i/2
a₂ (1 − i) / 2 ∨ a₃ = (1 + i) / 2
𝕃 = {1/2, (1−i)/2, (1+i)/2}
Thank you so much
Answer is there by just looking at it
Alright
Sufficient grace Chinwedu
Thank you so much sir
too straightforward
Ok
I did ...
a⁴ = (a - 1)⁴
a⁴ - (a - 1)⁴ = 0
(a²)² - ((a - 1)²)² = 0
(a² - (a - 1)²)·(a² + (a - 1)²) = 0
(a² - (a² - 2a + 1))·(a² + (a² - 2a + 1)) = 0
(a² - a² + 2a - 1)·(a² + a² - 2a + 1) = 0
(2a - 1)·(2a² - 2a + 1) = 0
/// case (2a - 1) = 0
2a - 1 = 0
a = 1/2
/// case (2a² - 2a + 1) = 0
2a² - 2a + 1 = 0
Δ = (-2)² - 4·2·1 = 4 - 8 = -4
√Δ = ±i√4 = 2i
• root #1: a = (-(-2) + 2i)/(2·2) = 2/4 + 2i/4 = 1/2 + i/2
• root #2: a = (-(-2) - 2i)/(2·2) = 2/4 + 2i/4 = 1/2 - i/2
/// final results:
■ a = 1/2
■ a = 1/2 + i/2
■ a = 1/2 - i/2
=> I T ' S A S Y O U R S O L U T I O N B E C A U S E I T ' S A V E R Y N I C E W A Y T O S O L V E T H E E Q U A T I O N .
🙂🙂🙂
Thank you so much for your kind words
And at the end of the day who really cares. And if it takes her that length of time to solve a fairly straightforward equation have all the students gone to sleep😂
😢
Ja, schon klar. Wurzel aus einer negativen Zahl.... Bei diesem i-Blödsinn steigt doch jeder normal denkende Mensch aus. Schon die Aufgabe selbst a = a - 1 zeigt doch, daß es hier keine vernünftige Lösung geben kann.
Okay