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

This is a nice Olympiad algebraic question. The solution was obtained using the laws of indices or exponentials. #matholympiadproblem #matholympiad #maths #matholympiadquestions #matholympiadpreparation #algebra

[@user-pk4nk1et9h]

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

[@JJONLINEMATHSCLASSchannel]

Yes, you are right. But the board is big enough, what I need is a better camera that will get the whole board. Thanks.

[@johnstanley5692]

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'.

[@JJONLINEMATHSCLASSchannel]

Thanks

[@prime423]

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!!

[@JJONLINEMATHSCLASSchannel]

Thanks but that will make the video to be too long

[@eldoserenado458]

Very good...

[@JJONLINEMATHSCLASSchannel]

Thanks

[@wilfredy.pasile7022]

Best maths teacher!

[@JJONLINEMATHSCLASSchannel]

Thanks for the compliment

[@BN-hy1nd]

Yes, you got me riveted. Well done 👍🏿

[@JJONLINEMATHSCLASSchannel]

I'm glad!

[@Honeyshaf]

To easy method for understanding the concepts

[@JJONLINEMATHSCLASSchannel]

Welcome

[@dbrovnievic]

You are so beautiful and so smart and on top of that, you have such a good-mood.

[@nasrullahhusnan2289]

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)

[@JJONLINEMATHSCLASSchannel]

Nice approach 🤝🤝

[@golddddus]

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😎

[@girmamumicha1642]

please teach us from basics on derivatives

[@JJONLINEMATHSCLASSchannel]

Ok. Noted.

[@MYeganeh100]

👌

[@JJONLINEMATHSCLASSchannel]

Thanks

[@wilfredy.pasile7022]

Genius lady indeed!

[@JJONLINEMATHSCLASSchannel]

Thanks

[@JomilHussainBarbhuiya]

Where are you from

[@JJONLINEMATHSCLASSchannel]

Nigeria

[@JomilHussainBarbhuiya]

Wow nice love from India

[@augustopinochet6841]

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

[@franklinbrown5625]

🇯🇲🇯🇲🇯🇲🇯🇲🇯🇲🇯🇲🇯🇲🇯🇲

[@JJONLINEMATHSCLASSchannel]

U are welcome