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1/2 dont ru n away

I just did it in my head

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Very good❤❤❤❤❤

Very good❤❤❤❤

ooga booga i ll clickbait them with "math olympiad" in title

I'm literally a math major, I have no idea why I clicked 😂

Nah sub for the biggest e term apply chen lu and get an answer lol did it within seconds

SO, WHAT'S THE X?

Shouldn't X be complexe ?

x=lnΦ/ln2 where Φ=(1+√5)/2 is the Golden Ratio.

these types of problems become very easy when you realize that the radical probably implies that what is under the radical probably takes the form of (a+b)²

In physics, similar algebraic form is given by the Fermi-Dirac distribution f(E) = 1/(1+exp((E-Ef)/KT)) .... and the identity f(Ef+E) + f(Ef-E) = 1 says that the the total probability of the conduction electron at the energy E above the Fermi level Ef and the probability of the valence electron at the energy E below the Fermi level is unity....

x * (x^2 + 1) ( x^3 + 1) = 0 x = 0, -1 are only feasible solution

x = 0, -1, +/-i, 1/2(1+/-sqrt(3)i). Going to watch video now to see if I am correct. OK x is real...so just 0, -1

I took a more intuitive approach. First, since all the signs were addition this means that the number can not be positive otherwise all the numbers would add up to more than zero. Since two of the Xs have even powers and the other two have odd powers this means that there will be two negative numbers and two positive numbers added together after carrying out exponentiation. The two odd powers will give negative numbers, the two even powers will give positive numbers. Since the powers of 6 and 4 are so much larger than 3 and 1 there is no number more negative than -1 that would cancel out all the terms to equal zero. The same is true for any number below -1 because the numbers cannot cancel out because of the big difference in powers being added. So the only number left is -1 which does work because of the way that it oscillates between positive and negative when going up by one in powers.

just a very complicated way to describe the golden section. Nice

I think your microphone setup could be improved. Have you tried a lapel mic?

I just looked at it and my guesses were 1, then 0, then 1/2. Feel like guessing is too strong here.

Man this is soo satisfying!

Very simple method of doing the problem. Nice

Another approach is to apply the difference of two squares.

Honestly, the root can be simply guessed by looking at the equasion close enough. Though for proving that the root is unique I've started thinking towards Lambert's W-function. Your method is much simpler)

since x is not equal to zero I would divide the whole equation by x^2 x^2+1/x^2 -7(x+1/x)+14=0 Let t=x+1/x then t^2= x^2+1/x^2 +2 Substituting t^2-2-7t+14=0 t^2-7t+12=0 t=3 or 4 Back substitute to get two quadratic equations of x and solve.

❤❤❤very good

I would use substitution van u=x+1 x -1 = u -2 x = u -1

Nice methode and video, just most of the time you put a one way symbol (-->) where it actualy in bothe wheys (<-->)

Why do you start by stating that x is real? X must be complex.

could you please as soon as possible set if you are looking for answers in natural numbers, real numbers or complex? For my undeerstanding, k is natural number, x is real and z is complex.

Maybe I'm missing something, but I think that there may be an error at 4:15 The equation is a^2 - 3xa + 2x^2 = 0 and it gets factored into (x - 2a)(x - a) = 0. But if I do the multiplication backward I get something that is not the initial equation: (x - 2a)(x - a) = x^2 - xa - 2xa + 2a^2 = 2a^2 - 3xa + x^2 I believe that the correct factorization should be (2x - a)(x - a)

like 53, buen ejercicio

Nice example

The methodology is (1) see the obvious solution (2) divide the cubic by (x - obvious solution) to give a quadratic (3) apply quadratic formula to get the other two solutions. RUclips is sadly full of "tutorial" videos that make step (2) look very mysterious by pretending that you haven't already done (1). Your method starting at 0:37 is not repeatable and doesn't work any better than mine when there isn't an obvious root.

Since the equation is symmetric divide the whole thing by x^2 then substitute t=x+1/x Much easier.

К = -2.

It's really easy tho. Looks very difficult.

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Why k=-2 is a solution? -12 is not equal to 12

Love the answer.

I do not understand the math at around 2:20 where you added things up

I know a cubic equation should have 1 real root. so I wanted get an intuitive guess on the real root first. Let f(k) = k^3 - k^2 + 12 this is clearly increasing for k > 0 as k^3 gets bigger than k^2 as k increases so real root must be where k < 0 let's search for root by substituting -1, -2, -3 .... we see that we got lucky at f(-2) = 0 so f(k) = (k+2)*(some quadratic) divide f(k) by (k+2) using algebraic long division f(k) = (k+2)*(k^2 - 3k + 6) k = (-3 +Sqrt(15)* i )/2, (-3 - Sqrt(15) * i )/2 and -2

U can directly substitute e^e^e^x as t.

the thumbnail says k^3 - K^2 = 12 and this is actually an interesting problem, the one here is trivial and can be solved in memory :)

Amazing explanation 😁

This can be easily done using cube roots of unity.

Multiply both sudes by 16. Substitute 2x+3 as r snd solve

(x+1)⁴ + (x+2)⁴ = 1 <=> (x+1)⁴ + (x+2)⁴ - 1 = 0 Substitute z := x + 3/2. Then let f(z) := (x+1)⁴ + (x+2)⁴ - 1 = (z - 1/2)⁴ + (z + 1/2)⁴ - 1. We know that f(z) is a polynomial of degree four, more specifically: f(z) = 2z⁴ + az³ + bz² + cz + 2(1/2)⁴ - 1, where a, b and c are real coefficients. Additionally, since (-z+1/2)⁴=(z-1/2)⁴ and (-z-1/2)⁴=(z+1/2)⁴, we can see that f(z) = f(-z). If we let z be complex, combined with the fact that f(1/2)=1-1=0 and f(-1/2)=1-1=0, we can conclude that there is a complex number d with f(d) = f(-d) = 0 (and there are no other roots since f is of degree four), such that we receive f(z) = 2(x-1/2)(x+1/2)(x-d)(x+d) = 2x⁴+bz²+cz+2(-1/2)(1/2)(-d)d= 2x⁴+bz²+cz+d²/2 With 2(1/2)⁴-1=d²/2 (see the above formula for f) follows d² = 2²(1/2)⁴-2=1/4-2=-7/4 Therefore, d = ±i sqrt(7)/2. Resubstituting, we then receive the roots x_1 = 1/2 - 3/2 = -1 x_2 = -1/2 - 3/2 = -2 x_3 = i sqrt(7)/2 - 3/2 = (i sqrt(7) - 3)/2 x_4 = -i sqrt(7)/2 - 3/2 = -(i sqrt(7) + 3)/2 Therefore, the only real solutions are -1 and -2.

Not too good at Olympiad integrals but got this one immediately!

By inspection x=-1 and -2 are roots of the given equation. Let y=x+1 and 2y^4+4y^3+6y^2+4y=0 or y^3+2y^2+3y+2=0 after dividing by 2y. Divide by (y+1): y^2+y+2=0 with roots (-1±i√7)/2 and x=(-3±√7)/2