Solving a septic equation 

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.

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

It is easy to guess the two solutions x= 0, x = -7 , but one has to show that these are the only real solutions.

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

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

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

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

Pascal Triangle 📐

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

( x + 7 )^7 - ( x^7 + 7^7 ) = 0 49 x ( x + 7 )( x^2 + 7 x + 49 )^2 = 0 x = { 0 , - 7 , 7 w , 7 w^2 } x^3 - 1 = ( x - 1 )( x^2 + x + 1 ) = 0 x = { 1 , w , w^2 } , w € C , w^3 = 1 😊🤪👍👋

(x+y)^7-x^7-y^7=7xy(x+y)(x^2+xy+y^2)^2　　; why (x^2+xy+y^2)^2　It's not a math formula, but there's no explanation.

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

(X+7)^7=X^7+7^7 X=-7 ,X=0,X=(-7±7Sqrt[3]i)/2=-3.5±3.5Sqrt[3]i

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

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.

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

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

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

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.

Alternative Solution for real roots only. Consider the function f(x)=(x+7)^7 - (x^7+7^7) First, compute the derivative: f'(x)=7(x+7)^6-7x^6 Setting the derivative to zero to find critical points: f'(x)=0 (x+7)^6=7x^6 Taking the sixth root on both sides: |x+7|=|x|. This implies x=-3.5, which is the only extremum and minimum point of the function. Since f(x) monotonic and continuous, it intersects the x-axis twice. Additionally, x=-3.5 is the axis of symmetry for the function derived from the binomial expansion. Given this symmetry, the second solution is 3.5 units away in the negative direction from the axis of symmetry at x=-3.5, which gives x=-7. Therefore, the real solutions are x=-7, x=0.

I don't like how you have to explain how to solve 49x=0 but not how to simplify (x+7)^7-x^7-7^7. Here's how I would have done it. Being able to cancel out the x^7 and constant terms is too good, so I would expand the polynomial. Don't want the coefficients blowing up? Substitute x=7t and the problem reduces to (t+1)^7=t^7-1. After the expansion, subtraction, and division by 7, we get t^6+3t^5+5t^4+5t^3+3t^2+t. Factor out the t, and factor the rest by grouping terms with the same coefficients. t^5+1+3t(t^3+1)+5t^2(t+1)=(t+1)(t^4-t^3+t^2-t+1+3t^3-3t^2+3t+5t^2)=(t+1)(t^4+2t^3+3t^2+2t+1)=(t+1)(t^2+t+1)^2. Solve for t, multiply by 7 to get x.

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 🤔

why not just expand using binomial and then cancel out the x⁷ and 7⁷ terms you can factor it out afterwards easily....

can you answer my questions i send to your email 🙏🙏🙏🙏it's secondary school problem and really need your help. Thanks

I generalised this for n is odd. Tried doing it for n is even and couldn't get anywhere 😩 Solve for x in terms of n if (x + n)^n = x^n + n^n and n ∈ Z^+. Case 1 - n = 1 : →x + n = x + n There are no valid solutions for x. Case 2 - n is odd and n ≥ 3: →(x + n)^n - x^n - n^n = 0 After looking at n = 3, 5, 7 and so on, we notice a pattern: →(n^2)*x*(x + n)*(x^2 + nx + n^2 )^((n - 3)/2) = 0 →x = 0, x = -n For x^2 + nx + n^2 = 0 , where n > 3: →x = (-n ± √(n^2 - 4n^2 ))/2 →x = (-n ± n√3*i)/2 If anyone can provide a generalisation for n is even, then please reply to my comment 😊

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

7(7¹x⁶ + 7⁶x¹) + 21(7²x⁵ + 7⁵x²) + 35(7³x⁴ + 7⁴x³) = 0 (7¹x¹)(x⁵ + 7⁵) + 3(7²x²)(x³ + 7³) + 5(7³x³)(x + 7) = 0 x[(x⁵ + 7⁵) + 3(7x)(x³ + 7³) + 5(7²x²)(x + 7)] = 0 *x = 0* (x⁵ + 7⁵) + 3(7x)(x³ + 7³) + 5(7x)²(x + 7) = 0 x³ + 7³ = (x + 7)³ - 3(7x)(x + 7) x⁵ + 7⁵ = (x + 7)⁵ - 5(7x)(x³ + 7³) - 10(7x)²(x + 7) x + 7 = a 7x = b (x + 7)⁵ - 5(7x)(x³ + 7³) - 10(7x)²(x + 7) + 3(7x)[(x + 7)³ - 3(7x)(x + 7)] + 5(7x)²(x + 7) = 0 a⁵ - 5b(x³ + 7³) - 10ab² + 3b(a³ - 3ab) + 5ab² = 0 a⁵ - 5b(a³ - 3ab) - 10ab² + 3b(a³ - 3ab) + 5ab² = 0 a⁵ - 5a³b + 15ab² - 10ab² + 3a³b - 9ab² + 5ab² = 0 a⁵ - 2a³b + ab² = 0 a(a⁴ - 2a²b + b²) = 0 a(a² - b)² = 0 a = x + 7 = 0 => *x = -7* a² = b => (x + 7)² = 7x x² + 14x + 49 = 7x x² + 7x + 49 = 0 x = (-7 ± 7i√3)/2 *x = (7/2)(-1 ± i√3)*

What is the simplification of (x+y)^n -x^n -y^n??

Would your experience solving this septic equation qualify you to repair our nasty, leaky, smelly septic tank? Nice job on choosing a relatively obscure term like septic as it could possibly enhance Search Engine Optimization(SEO), resulting in more page views from wordsmiths!

x = 0 ez

Bro u got to be the best Maths teacher

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?

Why couldn't we use the Pascal triangle for the first part (x+7)^7 ?

I have a septic infection 😂

For x in Z/7Z, we always have the equality : (x+7)^7 = x^7 + 7^7

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.

(x+7)^7=x^7+7^7 (0+7)^7=0^7+7^7 7^7=7^77=7 x=0

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.

Because x^2+7x+49 is squared can -x^2-7x-49 be used to solve for two other roots rather than repeat?

is there a general formula for factoring (x+y)^(2n-1) - x^(2n-1) - y^(2n-1)

My Ex was septic..

A septic equation turned into a sextic equation..... I never thought that algebra so "dirty".

why only six answers? shouldn't there be seven?

Fermat conjectures

Interesting to do the factoring, I'll try. But I'm curious how such things are obtained, I'd guess that can be done by synthetic division , if you have a clue what to obtain at tĥe end. Not quite obvious. Especially with the 7th degree, that incomplete square squared, looks overwhelming, I'd say 😅

septic… hm… 😂

The 7th root is ∞.

At the very end you say that 49 - 4(49) is negative 3 but it's negative 3(49) aka 147

I actually got the first and last term thing right, I just didnt know how to get the numbers in the middle lol

Eulers equation (a +b)^n= a^n+ b^n....for n=1,2....

Ответ один, а и 0 тоже...

I wouldve just said by fermas last theorem x can only be equal to 0

Don’t do it like this just brute force it and then synthetic Devine it Trust me man trust me

This is definetly an algebra's student dream.

Septic ?

isn't that equation more simple using pascal triangle ?

X=0 😂

multiplicity solution at end has been repeated

Medical person me reads “septic” 🤒

0

Its more like a hexic (is that the word for 6?) Rather than septic because the x⁷ terms cancel each other

Its true for all x in Z/7Z

Couldnt you just 7th root the entire equation and have all the exponents cancel out?

Mei muslman hon hindu nhi hon

put some TCP on it !!!😅😃

Tús es o cara. Thank you

To me its clear at the start that x must be less than 1.

Amazing

X=0

x=0

Believe it or not, I have made a summation for this exact problem but for all n not just 7

I really like this one!

Do me a favour: Don't call the non-real solutions 'imaginary'! They are called 'complex', ok. Nevertheless, the 'number' i ist called the 'imaginary entity'. Furthermore, there are 'imaginary numbers'. These are complex numbers without a real part or having zero as real part respectivly.

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