A nice math Olympiad problem |you should try to solve this
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- A nice math algebra simplification | you should know this trick
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There's a nice way using Viete's formulas - a,b are the roots of the equation x^2-(a+b)x + ab = 0, after finding ab=-1/2 you get 2x^2-2x-1 = 0 or x^(n+2)=x^(n+1) + (x^n)/2 - always true for roots a,b, giving the recursive relationship a^(n+2) + b^(n+2) = (a^(n+1) + b^(n+1)) + (a^n + b^n)/2.
b=1-a substitute for b in a^2+b^2=2. You will get 2*a^2-2*a-1=0
you can find a and b
Questo è il modo migliore, non quello del video, che non so dove porti.
Yeah, but it takes literally hours to raise it to the power of 11 and calculate🤦♂️🤦♂️
Let Sn be a^n + b^n Then you can show Sn+1 = Sn + 1/2Sn-1
S11 can be calculated with primary school math starting from S1 and S2
And S_0 = a^0 + b^0 too!
DUDE YOU NEED TO DRAW 1'S NOT LIKE STICKS CAUSE I THOUGHT YOU WERE TAKING THE DOUBLE DERIVATIVE OF A AND B!!!
Same😭
not for gymnasium students
YEAH! And in the answer it says g8g divided by 32. I never saw where the value of "g" was determined!!!
@@thomasharding1838are u stupid lol it’s 989
@@gilbertodeoliveirafrota5345hm? Not?
i ❤ Mathematics
Time taking and lengthy process. If you can find value of a-b by the formula (a-b)^2=(a+b)^2-4ab, after that find the value of a & b , finally you put the value get the answer
Perfect solution..👍. Here there is no other shortcut trick.. What is the point to find value of a and b?
The point is, that both solutions are equal ... technically each valid expression for "?" is a solution from math point of view.
RHS value = power of a or b= 11
Don't think too much
it is very easy to find the value of a-b
Long proces
Too long , attention span is 3 seconds max.🤣🤣🤣🧐🧐
1:07 too boring and complicated
Let a+b =S and ab=P. In the second equation a^2+b^2 =(a+b)^2-2ab=2 leads to ab=-1/2. If a and b are solutions of the quadratic equation X^2+PX+S=0, that give us to a,b =(1±√3)/2.
What I was thinking tbh
2
2 ans
I am always puzzled when I see solutions like that. Who the hell need it? You have two equations with two variables. Easily find both of them as (1+=sqrt(3)/2 and after any of their combination. How twisted mind should people have to post 16 minutes video with crazy computations for that task?
Now he got a result in Q where your result needs to be assumed to be in R ... unless you can get dont get rid of the "twisted mind" root in the term
(x+sqrt(y))^11 + (x-sqrt(y))^11 ... whereas with binomial sum at least the odd exponents for sqrt(y) should cancel and the even are integer then 🤔
@@zyklos229Too complicated. In this particular case quadratic equation has 2 real roots. No other complex roots exists. To raise them in power 11 is not rocket science...
ليه المرار الطافح ده ، استخدم نظرية ذات الحدين واخلص
Algebra hates it
Solutions
a=(1+3^1/2)/2
b=(1-3^1/2)/2
If you swap a and b, that is also a solution.
C’est vraiment passer derrière son genou pour se gratter la tête. 😂
2789/32 is result
1789/32.
“Fantastic explanation! I love how you break down complex concepts. I do similar math tutorials and challenges on my channel. Check it out for a fresh perspective on math problems!”
11 (??)
What is 13multiplied by13,you did not check also
169
You never said anything about 2ab, thus you changed the terms of the puzzle.
Plus you misspelled "equation" early on, and your 1's look horrid.
The answer is a=1.5 and b=-0.5. This fits the two terms of the puzzle that was stated.
Oh, and your "music' is repetitive.
Thank you for your feedback 😊
@@mks-learning-easy You are welcome! The best of luck to you!
Besides, to multiply equations, you bring all terms to one side (e.g. a+b-1)=0 and then multiply them. You don't just multiply left terms with left terms and right terms with right terms. In other words, your solution is horse shit.
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