Solving a septic equation

In this video, I solved a septic equation by considering a pattern of factors in the difference of polynomials of higher degrees

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[SpiroGirah]

Algebra is the king of mathematics. I wish I truly spent time developing that aspect of my math before calculus and other things showed up.

[mac_bomber]

10:39
"Those who stop learning, stop living"
Is that a threat?

[adwz]

Technically it’s a hexic (or sextic??) equation as the x^7 on both sides cancel, which means there should be 6 roots in C including multiplicity, as u found

[dougaugustine]

I watched this video twice because I like watching you solve problems like this.

[wavingbuddy]

Guys look at my cool millionth degree polynomial: x¹⁰⁰⁰⁰⁰⁰ = x¹⁰⁰⁰⁰⁰⁰ + x-1 😂

[kornelviktor]

The easy way to memorize 49 times 7 is 50 times 7 is 350 and minus 7 is 343

[pojuantsalo]

I suppose sanitary engineers need to solve septic equations...

[bobbun]

I would have to go check my old abstract algebra textbooks to find the exact way it's described, but if I remember correctly, for any prime p, you have (x+y)^p=x^p+y^p for all x and y when considering over the field Z_p. The trick is to realize that for a prime, all the binomial coefficients in the expansion of the left hand side are a multiple of p, except the first and last (which are always one). Since p~0 in that field, all the extra terms simply disappear. Granted, the solution over the complex numbers given here is the best interpretation of the problem when given without a more specific context, but it's nice to know that there really is a context where the naive student's thought that (x+y)^2=x^2 + y^2 actually does hold.

[trankiennang]

I think i have a general solution to this kind of equation: (x+n)^n = x^n + n^n ( n is natural number, n > 1).
Divide both side of equation by n^n. We will have (x+n)^n / n^n = x^n / n^n + 1 which is equivalent to (x/n + 1)^n = (x/n)^n + 1.
Let t = x/n, then the equation will become (t+1)^n = t^n + 1. So now we will focus on solving t
It is easy to see that if n is even then we just have one solution is t = 0 and if n is odd then t = -1 or t = 0. The main idea here is show that these are only solutions.
So let f(t) = (t+1)^n - t^n - 1
Case 1: n is even
f'(t) = n.(t+1)^(n-1) - n.t^(n-1)
f'(t) = 0 <=> (t+1)^(n-1) = t^(n-1)
Notice that n is even so n-1 is odd. Then we have t+1 = t (nonsense)
So f'(t) > 0. Thus f(t) = 0 has maximum one solution. And t = 0 is the only solution here.
Case 2: n is odd.
We have f''(t) = n(n-1).(t+1)^(n-2) - n(n-1).t^(n-2)
f"(t) = 0 <=> (t+1)^(n-2) = t^(n-2)
Notice that n is odd so n-2 is odd
Then we have t+1 = t (nonsense again)
So f"(t) > 0 which leads us to the fact that f(t) = 0 has maximum two solutions. And t = 0 and t = -1 are two solutions.
After we have solved for t, we can easily solve for x.

[mahinnazu]

Nice math solution.. I see you video everyday. It is really so helpful for me.
Thank you my Boss.
Mahin From Bangladesh.

[Blaqjaqshellaq]

The complex solutions can be presented as (7/2)*e^(i*2*pi/3) and (7/2)*e^(i*4*pi/3).

[timothybohdan]

Since the imaginary solutions get squared, you should also be able to use the negative of those imaginary solutions. Thus, the four imaginary solutions should be [+/-7 +/- i sqrt(3)]/2. Note the plus or minus in front of the 7. The other two (real) solutions are x = 0 and x = -7, as you noted.

[echandler]

Nice problem. Note that all of your solutions are multiples of 7: 0*7, -1*7, w*7 and (w^2)*7 where w and w^2 are complex cube roots of unity. This corresponds to your factorization.

[mitadas]

Can anyone please explain why the imaginary solutions are written twice?

[sajuvasu]

U can say complex solutions....
Anyway very informative 😁😁

[zyklos]

I would say (x + 7)^7 = sum i=0..7 binomial (7 over i) x^i 7^(7-i)
Leaving us with 1 x=0 solution and polynomial of degree 5 equals 0, so 5 more solutions, none of it positive. (-7) seems a solution, dividing leaves us with solveable 4th degree.
The shown factorization makes sense, but appears little bit abitrary 🤔

[maburwanemokoena]

This is definetly an algebra's student dream.

[donwald]

The only septic I can solve is figuring out what happens when I flush my toilet lol.

[ElAleXeX]

Could this mean we can express (x+y)^n as x^n + nxy(x+y)(x²+xy+y²)^((x-3)/2) + y^n where n is an odd positive integer?

[maharorand]

That s a rly cool explanation but the third is wrong to me : if ( x2 + 7x + 49 )2 equals 0 then x2 + 7x + 49 equals square root of 0 so 0 and x2 + 7x + 49 is ( x + 7 )2 so we replace and then we take out the square of ( x + 7 )2 so x + 7 = 0 and we get the same answer than the last one