Japanese | A Nice math Olympiad algebra problem | Solve for a and b. 

Accelerated Girard-Newton method: x^2-(a+b)x+ab=0 (Vieta). x^2-x-1/2=0 x^2=x+1/2 (1) S(k)=a^k+b^k (1)*x x^3=x^2+x/2 ⇒ S(3)=S(2)+S(1)/2=2+1/2=5/2. (1)*x^2 x^4=x^3+(x^2)/2 ⇒ S(4)=S(3)+S(2)/2=5/2+2/2= 7/2 (1)*x^3 x^5=x^4+(x^3)/2 ⇒ S(5)=S(4)+S(3)/2=7/2+5/4=19/4 (1)*x^4 x^6=x^5+(x^4)/2 ⇒ S(6)=S(5)+S(4)/2=19/4+7/4=13/2 Acceleration is your solution, direct multiplication, instead of boring calculation S(7), S(8), S(9) S(10). By the way x^2-x+abx^0 =0 ⇒ S(2)-S(1)+ abS(0)=0 S(0)=a^0+b^0=2 ab=-1/2😎

Could also use synthetic division: let g1=a + b - 1 (=0), g2=a^2 + b^2 - 2 (=0), g3= a^11 + b^11 =(?) 1st obtain p2(b) =remainder= (g2/g1) = 2b^2 - 2^b - 1 = 0 . (here use 'a' as independent to return function of 'b') 2nd step get remainder p10(b)= g3/g1. Only need coefficients: [11 -55 165 -330 462 -462 330 -165 55 -11 1] final step, compute remainder p10(b)/p2(b)= 989/32. Advantage? never need to obtain values of 'a' or 'b'.

Very clear process but you need a bigger board so that you keep most of the work available.

Problem solving outline given: a+b=1 and a²+b²=2 Outline: Find ab through (a+b)²=a²+b²+2ab Find a³+b³ through (a+b)³=a³+b³+3ab(a+b) Find a⁹+b⁹ through (x³+y³)³ Find a¹¹+b¹¹ though (x⁹+y⁹)(x²+y²)

Very good...

Best maths teacher!

One can use other exponentials to get the result. It just takes more time. Its important to show students other possible ways to get the same result. For example, the sum of a exponent 2 +b exponent 2 to the fourth power!!

To easy method for understanding the concepts

👌

Yes, you got me riveted. Well done 👍🏿

please teach us from basics on derivatives

Genius lady indeed!

If it is allowed to use the calculator, it can be solved within 5 minutes.

Where are you from

🇯🇲🇯🇲🇯🇲🇯🇲🇯🇲🇯🇲🇯🇲🇯🇲