X= -1.

ME IMAGINO A ESTA PERSONA EXPLICANDO LA TEORIA DE LA RELATIVIDAD!!!!!

Teacher thank u

Keep it up sir 🙌

X5=1_____therefore x=1

Bonne solution à Mr..

ok but some suggesting shorter easier method

Good morning, Sir. Please I have a question. How do I solve for n in (n³-3n²+2n-210=0) Sir, please if it possible can you do a video to explain it?

Please what is your take on this post

you make it look difficult, you can finish this problem in 3 steps

I was a better mathematician, if i had a helmet

wish this was in my olympiad 😢

Please sir can I join you

good video

Seriously? 4^7 * 11... Thats it

А зачем логарифмировать по основанию 10

Take 4⁷ common and you get 4²-4-1 in bracket now in side you get 11 and outside you get 4⁷. ?

A quicker method is to take out a common factor of 4^7 first, then you have 4^7 x (16-4-1) = 16384 x 11

Appreciate your appreciation of safety

This is not trigonometry,this is geometry 😅

Зачем все эти извращения (да еще и с переписыванием двух строк два раза туда-сюда) если можно в исходное уравнение вместо Х подставить Х-3, возвести в квадрат, раскрыть скобки и получить то же самое,

Which Olympiad is this from?

Brilliant process. Keep it up

1 and 4 complex roots.

This is very good

Like the hat😊

Good work ✅

Why can't you take the fourth root from the beginning?

Please what is your take on this solution

🥶 ☃️

F(x)=F(x+3)+6x Hhhhhh😅😅😅😅😅😅

Awesome❤

شاباش۔

@ wisfrank -- I suggest you try a different factorization: x^5 + x^4 + x^3 + x^2 + x + 1 = 0 x^3(x^2 + x + 1) + 1(x^2 + x + 1) = 0 (x^2 + x + 1)(x^3 + 1) = 0 (x^2 + x + 1)(x + 1)(x^2 - x + 1) = 0 Continue solving as you did.

x²-3x+2.

i am not understanding

Nice solution 👍

There was no need for such complications. The value of f(x) is easily obtained by substituting x-3 for x. f(x-3+3)= (x-3)^2 + 3(x-3)+2 ; i e f(x)= x^2+9-6x+3x-9+2 ; i e f(x)= x^2-3x+2.

Please can I join?

X^5 + X^4 + X^3 + X^2 + X + 1 = 0 easy to see X=-1 is a solution. divide trough ( X +1 ) gives X^5 + X^4 + X^3 + X^2 + X + 1 = (X+1)(X^4 +X^2 + 1 )= 0 thus X = -1 or X^4 + X^2 + 1 = 0 this give no real zero s. X^2 = (-1/2) +/- (1/2)(3)^(1/2) * i gives four complex solutions. thus the only real solution is X = -1

Thank you for the insight! Keep pushing!!

👏🏼👏🏼👏🏼

Brilliant!

I basically took only the real answers, but here is how to solve it in only 4 lines: (x-3)⁴=16 (x-3)⁴=2⁴ x-3=2 and x-3=-2 x=5. x=1 There are two solutions for this equation.

Thanks

9 is not a prime number, for 9=3*3. Nor is 57 a prime number, since 57=3*19. Therefore the factorization should be 3*3*3*7*19*73.

Use x^6=1+(x-1)*(1+x+x^2+x^3+x^4+x^5);given(1+x+..+x^5 = 0) , (x~=1), gives x^6= 1 =exp(i*2*n*pi) => x=exp(i*n*pi/3), n=1,2,3,4,5 this is easier way to obtain your result. Also note this identical problem has been posted on KZbin many times.

Good job

Good job 👍

f(x)=f((x-3)+3)