Way too complicated. I'll show you how I got x = 2 in my head. No estimations (okay, honestly a bit of trial and error), only using the Pascal's Triangle: So, the binomial theorem works by using the Pascal's triangle to get the coefficients. (4 + sqrt(15))⁴ is equal to 1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 -> this row So: 1 • 4⁴ • sqrt(15)⁰ + 4 • 4³ • sqrt(15)¹ + ... However, this property changes for (a-b)^n in the aspect that every second coefficient will become negative. Meaning that the result above will become: 1 • 4⁴ • sqrt(15)⁰ - 4 • 4³ • sqrt(15)¹ + 6 • 4² • sqrt(15)² - 4 • 4¹ • sqrt(15)³ + 1 • 4⁰ • sqrt(15)⁴ The pattern you can see here is that every coefficient, where sqrt(15)^n in which n ł 2, will cancel out by addition above. This already shows us that the answer x is highly likely to be an integer because we get rid of all the roots. After then, I tried out: 4⁴ > 62, doesn't work 4³ = 64 > 62, doesn't work 4² = 16, that might be possible. so since the binomial coefficients for x = 2 are 1 2 1 and we know that every coefficient with sqrt(15)^n in which n ł 2 is canceled out, which is the coefficient 2, so 2 • 4¹ • sqrt(15)¹ this leaves us with 4² + sqrt(15)² = 31. double that, because we are adding (a+b)^x and (a-b)^x and our result is 62, immediately concluding that x = 2 is our solution.

Risolvo la quartica..2+2x^4=1+6x+4x^2+6x^3+x^4..dopo i calcoli risulta (x^2-3x+1)^2-15x^2=0..poi è semplice .

At the beginning of the 2nd page (~5:18), you know x^4 - 4x^3 - 6x^2 - 4x + 1 = 0 Observe that flipping the sign of 6x^2 gives you (1-x)^4, so we can in fact rewrite the equation as (x^4 - 4x^3 + 6x^2 - 4x + 1) - 12x^2 = 0 i.e. (1-x)^4 - 12x^2 = 0 This is the difference of two squares, factorize it and you would get two quadratic equations, which give you the 4 solutions of x. It would save you some times compare with the t substitution.

At the end you say all 4 solutions are real not complex, however it is not true. x_3 and x_4 are complex, as [3 - 2*sqrt(3)] is, in fact, negative.

Very beautiful question Thank you professor 🙏🙏💵💴💵🎩

Pour x3,x4, c'est 2 nombres complexes car 3-2racine3 est négatif

A very nice solution indeed.

A=(4+15^1/2 )^x , B =(4- 15^1/2 )^x. Then AB =1, A+B=62. Therefore A, B are two solutions of x^2 -62x +1=0.

You several times said "irrational number" when you meant "not the square of an integer". I hope you correct your script and rerecord this.

0 is not natural number

Moanus

Great job 👏

If you divide by 2 5 times you get an odd number which tells you that 2^{a-5}=1 or 2^{b-5}=1. This means either a=5 or b=5. Computing 2^5=32 makes it obvious that a=5 cannot work as 32<2016 and subtracting 2^b for any b will only make the expression smaller. Thus, b=5 from which we can compute 2^a=2016+32=2048 At this point, you can divide by 2 until you compute a, but another trick to remember (especially if you are a computer scientist) is that 2^10~1000, from which we gather that a=11 is the only possible solution

/:x^ln4 , 1+x^(ln10-ln4)=x^(ln25-ln4) , 1+x^(ln10/4)=x^(ln25/4) , 1+x^(ln5/2)=x^(ln25/4) , (5/2)^2=25/4 , x^(ln25/4)=2*(ln5/2) , 1+x^(ln5/2)=x^(2*ln5/2) , let u=x^(ln5/2) , u^2-u-1=0 , u=(1+/-V(1+4))/2 , u= (1+V5)/2 , / (1-V5)/2 < 0 , not a solu / , x^(ln5/2)=(1+V5)/2 , lnx=ln((1+V5)/2)/(ln5/2) , x=e^(ln((1+V5)/2)/(ln5/2)) , x=~ 1.69045 , test , x^ln4 +x^ln10 = 5.42202 , x^ln25 = 5.42202 , OK ,

X,Y natural ədədlər çoxluğuna daxil olduğundan X+Y>X-Y. Təşəkkürlər.

I just used the binomial theorem and evaluated the cubic for real roots and then found the other roots using the quadratic formula, took me 4 minutes without pen and paper needed

2^a - 2^b. = 2⁵ . 63 On simplifie par 2⁵ 2^(a-5) - 2^(b-5). = 63 = 64 - 1 =. 2⁶ - 2⁰ D' où a-5 = 6, donc a = 11 b-5 = 0, donc b = 5

x = 0 ===>> 1 + 1 = 62 (F) x = 1 ===> 4 + 4 = 62 (F) x = 2 ===> 30 + 32 = 62 (V) S = {2}

6^(a+b) = 12 * 18 = 6^3 => a + b = 3 Notice that 3/2 = 18/12 = 3^(a - b) / 2^(a - b) = (3/2)^(a - b) => a - b = 1 Thus, adding the equations a + b = 3 and a - b = 1, we get 2a = 4, or a = 2. This also tells us that b = 1. Thus, a = 2 and b = 1, so (a + b)^(ab) = 3^2 = 9.

X=2 put here and answer solved 😂

Merci beaucoup , j'ai pu résoudre correctement cet exercise avant de regarder la vidéo

((4+V15)/(4-V15))=(4+V15)*(4+V15) , 1/(4-V15)=(4+V15) , let u=(4+V15)^x , u^2+1=62u , u^2-62u+1=0 , u=(62+/-V(62^2-4))/2 , u=(62+/-V(3840))/2 , u=31+/-V960 , u=31+/-8*V15 , / 31-8*V15 > 0 , OK / , case1 , (4+V15)^x=(31+8*V15) , x=log(31+8*V15)/log(4+V15) , x = 2 , case 2 , (4+V15)^x=(31-8*V15) , x=log(31-8*V15)/log(4+V15) , x= -2 , test , x=2 , (4+V15)^2+(4-V15)^2=62 , OK , test , x=- 2 , (4+V15)^(-2)+(4-V15)^(-2)=62 , OK , solu , x = 2 , -2 ,

By hit and trial put x=2 it will be (a+b)^2 + (a-b)^2 which will be 2(a^2+b^2) which is 2*(4^2+√15^ 2)= 2*(16+15)= 2*31= 62 which statisfies so x=2

x=2, 4 work, lol...

Ans is 2^c

Solution: 27^[lg(x)]+3^[lg(x)] = 68 ⟹ 3^[3*lg(x)]+3^[lg(x)] = 68 |with u = 3^[lg(x)] ⟹ u³+u = 68 = 4³+4 | The same operations are done with u on the left side of the equation as with 4 on the right side of the equation, therefore: u = 4 = 3^[lg(x)] |lg() ⟹ lg(4) = lg(x)*lg(3) |/lg(3) ⟹ lg(4)/lg(3) = lg(x) |10^() ⟹ x = 10^[lg(4)/lg(3)] = 10^[lg(4)]^[1/lg(3)] = 4^[1/lg(3)] ≈ 18,2751

Isn't x = 2?

8^(x)-2^(x)/3^(x)*(2-1)=2 8^(x)-2^(x)/3^(x)=2(2-1) {[8/3]^(x)} - {[2/3]^(x)=2 [6/3]^(x)=2 2^(x)=2 x*ln(2)=ln(2) x = 1.

Can't understand why such complicated solutions are presented. Multiply both sides with (6^x - 3^x). You get 8^x-2^x = 12^x-6^x The only solution is x=1. x=0 is excluded: Reason is, that division by 0 is not allowed. x=2 would have 20/9 = 2.22222 as result of the initial equation.

Why not approach it like this: The square roots of -64 are 8i and -8i. Then, we can take the square roots of one of these, and the other will simply come from flipping the sign of one number or the other. In this case, (2+2i)²=4+4i+4i-4=8i. So we get, as required, four fourth roots of -64: 2+2i, -2-2i, 2-2i, and -2+2i.

x=-6

1) Expand both binomial in Newton bonimal 2) Note that most of them cancel each other OBS: Consider x natural and test for x = {0,1, 2, 3, 4, ...} At least i tried 🙂

what is the value of W function in the last step? It is like a problem in itself

Is W(b) calculated as the a such that ae^a = b?

I looked at the problem for less than 5 seconds and tried it with x=1. Then the equation becomes (8 - 2 ) / (6 - 3) or 6 / 3 which is 2. No need to waste nearly 14 minutes on that.

At the the qusrtic stsge z& Z^4 could have been replsced by k^2 (k =z^2).

(x+9)⁴ + (x+11)⁴ = 706 ... let u = x+10 ... (u-1)⁴ + (u+1)⁴ = 706 (u⁴ - 4u³ + 6u² - 4u + 1) + (u⁴ + 4u³ + 6u² + 4u + 1) = 706 2u⁴ + 12u² + 2 = 706 2u⁴ + 12u² = 704 u⁴ + 6u² = 352 u⁴ + 6u² + 9 = 361 (u² + 3)² = 19² u² + 3 = ±19 u² = -3 + 19 OR u² = -3 - 19 u² = 16 OR u² = -22 u = ±4 OR u = ±i*√22 x+10 = -4 OR x+10 = +4 OR x+10 = -i*√22 OR x+10 = +i*√22 x = -14 OR x = -6 OR x = (-10 - i*√22) OR x = (-10 + i*√2)

Решаем методом устного счета. х=1 8-2=6 6-3=3 6:3=2. Good luck!

x^2+6x+4=a buradan x^2+6x=a-4 (a-4)(a+4)=9 a^2-16=9 a^2=9+16=25 , a=5, a=-5 x^2+6x+4=5 x^2+6x+4=-5 x^2+6x-1=0 x^2+6x+9=0 x=-3+-*(-3)^2-(-1) (x+3)^2=0 x=-3+-*9+1 x+3=0 x=-3+-*10 x=-3 x=-3-*10 x=-3+*10 Təşəkkürlər.

I see. This problem can be solved using Calculus 1 theorem of Lagrange. Let f(x) be a function continuous on definition interval [a,b] and derivable on interval (a,b) . Then according to graph there exists a point c €(a,b) such that f'(c)*(b-a)=f(a)-f(b). The graphical interpretation is that there exists at least a tangent to the graph parallel to the line passing trough thw atartpoint and endpoint. We notice that 8^x-2^x=2(6^x-3^x) looks similar to theorem of Lagrange. Let f(t)=t^x be a polynomial function. As a polynomial it satisfies the continuity and derivavility requirements so we can apply for intervals [2,8] and [3,6] to obtain: There are pints c in d so that f'(c)(8-2)=8^x-2^x <=> 6f'(c)=8^x-2^x And d such that f'(d)(6-3)=6^x-3^x <=> 3f'(d)= 6^x-3^x The original expression becomes 6*f'(c)=2*3*f'(d) But f'(t)=x*t^(x-1) so the equation becomes X(c^(x-1)-d^(x-1))=0 Since c and d belong to overlapping intervals they might be the same power which would mean any x satisfies the condition but if we plug in 2 for example we notice this is not true so c and d are different so the two exponentials only intersect when x=1. We can verify that x=0 is not a solution as denominator is 0 This approach not only let us know X=1 is a solution but also it is the only solution Edit: I didn't see that we are only looking for integer solutions from the splash screen

>olimpiad problems >okay

Can't understand what has being numbers odd and even anything to do with equating sides.

f((2x-1)/(x-3)) = x² Let u = (2x-1)/(x-3) . u(x-3) = 2x-1 ux - 3u = 2x - 1 ux - 2x = 3u - 1 x(u-2) = (3u-1) x = (3u-1)/(u-2) x² = (3u-1)²/(u-2)² ==> f(u) = (3u-1)²/(u-2)²

Its 2. there, done.

At 03:30 there seems you missed 1 square x. Am I wrong?

You can use a substitution b = a - 1/2 Then: (b-1/2)^4 = (b+1/2)^4 Developing the equation all even power coefficient cancels out and you get: 4b^3+b=0 b(4b^2+1)=0 Which can be easily solved obtaining b = 0 or i/2 or -i/2 Using a=b+1/2 you obtain the solutions

What is the range for which the function is defined?

The start of the video reminds me of finding the inverse

k(k^2+1) = 2*5 k^2+1 = 5 k = 2

Why not square root both sides? If the initial value is equal then their square root must be equal too.