Solving A Quartic Equation | Problem 349

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Опубликовано:

18 сен 2024

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Комментарии : 26

@mcwulf25 13 дней назад

I spotted #3. But we don't need to divide at the start. Just multiply by 1 then take 6th roots.

@aplusbi 13 дней назад

Good thinking

@leonidfedyakov366 13 дней назад

oh, that Syberian accent is so dear to my heart )))

@aplusbi 12 дней назад

😃

@Nobodyman181 13 дней назад

Hexxic

@YoursTrulyAkr 12 дней назад

"Two C or not To see " I wonder how many people end up understanding the awesome jokes you Crack during your vids Loved it

@aplusbi 10 дней назад

Thank you!

@caseyglick5957 11 дней назад

That final expression can be "simplified" to be a function of the cotangent. Distribute exp(pi n / 6) out of top and bottom, and you get i cos(pi n / 6) / sin(pi n / 6). Again, you can't do n=0, but we know what the various other options are. You get two roots of at n= +/- 3, where z=0, and then the other two roots at n = +/- 1/6 and +/- 1/3

@caseyglick5957 11 дней назад

sin(pi/6) = 1/2 and cos(pi/6) = sqrt(3)/2 (and the converse for pi/3). This gives the roots of 1/sqrt(3) and sqrt(3).

@scottleung9587 13 дней назад

The binomial theorem was the first thing that came to mind for me.

@trojanleo123 13 дней назад

z = 0; z = ±√3i; z = z = ±i/√3

@leonidfedyakov366 11 дней назад

Let’s multiply a+bi by sexual. We have asexual + bisexual. This is truly complex and more than a number.

@mapwiz-sf5yt 11 дней назад

Probably worth pointing out that z=0 is a double root.

@Nobodyman181 13 дней назад

(z+1)⁶-(z-1)⁶=0 ((z+1)³+(z-1)³)((z+1)³-(z-1)³)=0 1)(z+1)³+(z-1)³=0 (z+1+z-1)((z+1)²-(z+1)(z-1)+(z-1)²=0(using formula for a³+b³) 2z(z²+2z+1-z²+1+z²-2z+1)=0 z(z²+3)=0 z=0 or z=±√3i 2)(z+1)³-(z-1)³=0 (z+1-z+1)((z+1)²+(z+1)(z-1)+(z-1)²)=0(using formula for a³-b³) z²+2z+1+z²-1+z²-2z+1=0 3z²+1=0 3z²=-1 z=±(1/√3)i

@richardmullins44 10 дней назад

z=0 . (I guess there are other answers too). I checked with open ai, it did not find any other answers. A comment - I think the people posting these problems are usually overthinking the problem. These problems are meant to be answered at a glance, so that you can spend more time on problems that are more difficult. If you know that 1^6 = (-1) ^ 6, then you can see that z = 0 is an answer.

@neuralwarp 13 дней назад

(z+1)⁶ = (z-1)⁶ z+1 = z-1 +1 = -1 QED

@aplusbi 13 дней назад

🤪

@kappasphere 10 дней назад

I actually did this, but I replaced z-1 with 1-z because x⁶ is symmetric. That at least gives you one of the solutions, but doesn't get you further than that

@nowymail 12 дней назад

0. Solved in 5 seconds.

@kappasphere 10 дней назад

How many seconds did it take you to notice that it has 4 other solutions

@richardmullins44 10 дней назад

@@kappasphere Well done!! I asked open ai and it only found the answer 0. But after I saw your comment, I asked open ai if there were other solutions, and it found 4 more. It is probably fair to note that this question is given in an exam of short answer questions, and you are not at the very highest level of maths (either at high school or university), they probably only want the answer 0.

@leonidfedyakov366 13 дней назад

Solving hexic gets hectic. If we apply polynomial coefficients to the equation opening up the parentheses, we will end up with a quinary equation, as z^6 cancels out. Thus we have 5 roots of the equation, not 6.

@aplusbi 10 дней назад

I think it turns into a quartic

@phill3986 13 дней назад

☮️👍✌️😃✌️👍☮️

@key_board_x 10 дней назад

(z + 1)⁶ = (z - 1)⁶ (z + 1)⁶ - (z - 1)⁶ = 0 [(z + 1)²]³ - [(z - 1)²]³ = 0 → recall: a³ - b³ = (a - b).(a² + ab + b²) [(z + 1)² - (z - 1)²].[{(z + 1)²}² + {(z + 1)².(z - 1)²} + { (z - 1)² }² ] = 0 [(z + 1)² - (z - 1)²].[{(z + 1)²}² + (z + 1)².(z - 1)² + {(z - 1)²}²] = 0 → recall: a³ - b³ = (a - b).(a² + ab + b²) [(z + 1) + (z - 1)].[(z + 1) - (z - 1)].[{(z + 1)²}² + (z + 1)².(z - 1)² + {(z - 1)²}²] = 0 [z + 1 + z - 1].[z + 1 - z + 1].[{z² + 2z + 1}² + (z² + 2z + 1).(z² - 2z + 1) + {z² - 2z + 1}²] = 0 [2z].[2].[{z² + 2z + 1}² + (z² + 2z + 1).(z² - 2z + 1) + {z² - 2z + 1}²] = 0 4z.[{z⁴ + 2z³ + z² + 2z³ + 4z² + 2z + z² + 2z + 1} + (z⁴ - 2z³ + z² + 2z³ - 4z² + 2z + z² - 2z + 1) + {z⁴ - 2z³ + z² - 2z³ + 4z² - 2z + z² - 2z + 1}] = 0 4z.[{z⁴ + 4z³ + 6z² + 4z + 1} + (z⁴ - 2z² + 1) + {z⁴ - 4z³ + 6z² - 4z + 1}] = 0 z.[z⁴ + 4z³ + 6z² + 4z + 1 + z⁴ - 2z² + 1 + z⁴ - 4z³ + 6z² - 4z + 1] = 0 z.[3z⁴ + 10z² + 3] = 0 First solution: z = 0 Second case: [3z⁴ + 10z² + 3] = 0 3z⁴ + 10z² + 3 = 0 z⁴ + (10/3).z² + 1 = 0 → let: Z = z² Z² + (10/3).Z + 1 = 0 Δ = (10/3)² - 4 = (100/9) - (36/9) = 64/9 = (8/3)² Z = [- (10/3) ± (8/3)]/2 Z = - (5/3) ± (4/3) Z = (- 5 ± 4)/3 First case: Z = (- 5 - 4)/3 Z = - 3 Z = 3i² → recall: Z² = z z = ± i√3 Second case: Z = (- 5 + 4)/3 Z = - 1/3 Z = (1/3).i² → recall: Z² = z z = ± i/√3 z = ± (i√3)/3 Resume: z = 0 z = i√3 z = - i√3 z = (i√3)/3 z = - (i√3)/3

@aplusbi 10 дней назад

Wow!