To avoid such difficult calculation we can use concept of nth root of unity which says that if x^n=1 then x=e^(2πki/n) where k=0,1,2,3......(n-1) If you put k gretaer than that then roots will starts repeating and also we can simplify them in much nicer form by euler identity which says e^(ix)=cosx+isinx
The only integer solution is 1, and the others are going to occur in pairs as complex numbers. Whenever you have x^n + .... and you're solving, since polynomials split completely over C, you will get n solutions. An even number of them will be complex, and the rest of them will be real.
x = 1 is one solution. Polynomials split completely over C, so there are guaranteed to be five solutions, and since 5 is prime, they will all be unique, and four of them will be complex, occurring in two pairs.
At a glance, x needs to be 1 to satisfy the equation,
To avoid such difficult calculation we can use concept of nth root of unity which says that if x^n=1 then x=e^(2πki/n) where k=0,1,2,3......(n-1) If you put k gretaer than that then roots will starts repeating and also we can simplify them in much nicer form by euler identity which says e^(ix)=cosx+isinx
Yes, this method is unintuitive and very long winded.
√9=±3 and √8=±2√2....so there are complex roots too
this is not math. it's something else 😂😂
Luckily for you this is a needlessly complicated method. Just use the nth roots of unity
When i find something out of bound, i hit Thumb down button. And everything makes sense then.
x = 1 and very likley solutions with i .....
The only integer solution is 1, and the others are going to occur in pairs as complex numbers.
Whenever you have x^n + .... and you're solving, since polynomials split completely over C, you will get n solutions. An even number of them will be complex, and the rest of them will be real.
Racine niéme de l’unité. Réponse immédiate sans calcul.
X^5=1=1^5X=1
x = 1 is one solution. Polynomials split completely over C, so there are guaranteed to be five solutions, and since 5 is prime, they will all be unique, and four of them will be complex, occurring in two pairs.
Sorry. I will not comment.
Nul .... nul.
√9=±3 and √8=±2√2....so there are complex roots too
√9=±3 and √8=±2√2....so there are complex roots too
√9=±3 and √8=±2√2....so there are complex roots too