Another approach - first we find ab like in video. Then make quadratic with a,b as roots. Hence, x²-x-1/2=0 is our quadratic. Now let the sequence t_n be defined as t_n = a^n + b^n Now we know that a²= a+1/2 Multiplying a^n both sides a^(n+2)= a^(n+1) + (1/2)* a^n Similarly b^(n+2)= b^(n+1) + (1/2)* b^n Adding these we get t_(n+2)= t_(n+1) + (1/2)* t_n We know t_1= 1 and t_2=2. Using the above formula we can calculate t_3,t_4 and so on uptill the required t_11. For example t_3= t_2+ (1/2)*t_1= 5/2 t_4= 5/2+ (1/2)*2= 7/2 t_5= 7/2+ (1/2)*(5/2) = 19/4 and so on...
Sir i paused video and did it with different approach. With a+b and ab i made a quadratic equation then i found a and b then a¹¹+b¹¹ became a binomial expression and as a and b were conjugate so all odd terms got cut and remaining was the answer. I got the right answer.
@@Beyondfutura see a+b = 1 and ab=-1/2 Using this we can for a quadratic equation , 2a²-2a-1=0 then value of a = (1+rt3)/2 and b= (1-rt3)/2 Now a¹¹ + b¹¹ = [(1+rt3)¹¹+(1-rt3)¹¹]/2¹¹ now use concepts of binomial theorem.
You can use a different method for this using Viete's formulas. After calculating ab = -1/2, you can say that a,b are the roots of the quadratic 2x^2-2x-1=0. They are also the roots of 2x^11-2x^10-x^9=0, multiplying across by x^9, and 2x^10-2x^9-x^8=0. Adding these equations we eliminate the x^10 term to get 2x^11-3x^9-x^8=0 By repeatedly changing the degree of the leading coefficient and adding multiples to clear the second highest term, and clearing denominators, you can find that a,b are also the roots for the following polynomials: 4x^11 - 8x^8 - 3x^7 = 0 4x^11 - 11x^7 - 4x^6 = 0 8x^11 - 30x^6 - 11x^5 = 0 8x^11 - 41x^5 - 15x^4 = 0 16x^11 - 112x^4 - 41x^3 = 0 16x^11 - 153x^3 - 56x^2 = 0 32x^11 - 418x^2 - 153x = 0 Here's where the magic happens. Setting x=a,x=b, and added the expressions together, you get: 32(a^11+b^11) - 418(a^2+b^2) - 153(a+b) = 0 which gives a^11+b^11=989/32
Sir please dont stop making these type of videos where you pick an intresting question,i literally wait for them as i love math and love to be challenged by it😅
10:10 Mujhe (a^11+b^11) generate karne ka ah...hh....... yaha pe ek ahhh...hh........... infinite years later... Generate ho skati hai..😂😂 By the way thanks for this question i solved by another way sir 😍😍
I had a different approach, Prepare a quadratic equation using the given info. using Viete's formulae, The expected equation is, 2x^2 - 2x - 1 = 0 since, a and b are roots, define : S(n) = a^n + b^n then : 2S(n) = 2S(n-1) + S(n-2) S(n) = S(n-1) + (1/2)*S(n-2) a^n + b^n = S(n-1) + (1/2)*S(n-2) set n = 11, a^11 + b^11 = S(10) + (1/2)*S(9) This is an application of Newton's formula. After expanding the binomial, odd terms get cancelled and the answer is found to be (989/32).
Sir a aur b ke bich ke relation se unke liye solve Kiya toh ek answer a = -W aur b = -W^2 aata hai ... Where W is cube root of unity Waha se required value -1 aati hai
Initially I was also going in the same direction but I thought it would long process there might be some Quick way to approach this problem but I was wrong
Good and highly effective steps in solving this highly complex math problem in terms of powers.. ,the question of curiosity is what is the value of (a) and (b) respectively ..... thanks...
It's not complex, someone stupid as me with basic acknowledge about algebra could easily resolve it by using a system of equations. And the value of a and b aren't just one number, they has more roots, specifically, 2 roots each one, a (sub.1)=b (sub.2)≈-0.365 and b (sub.1)=a (sub.2)≈1.365.
The base of a number means frequency in physics. So why are there limits to frequency because they can combine only in specific patterns. Essentially time dimensions are bases. Structural coordination is f mod. Right angles mean that. Light can be modulated only in specifics of right angles. Energy is a constituent of absolute squares.
any one understood, this ana system of calculation of recursive values since, we need to half of last but one term, every time, after we get a term, we half it immediately for future addition operaiton
Since a+b = 1, you know that b = 1-A. Replace b with 1-A in the second equation. A^2 + (1-A)^2 = 2 Then solve with quadratic formula to find A. Then you have B = 1-A. Then calculate A^11 + B^11
Sir kya main aapse personal me baat kar sakta hu ? Kyuki main math me relativity dipole se related ek equation banana chahta hu ye dipole thoda complicated hai aur maine ise banaya hai par is dipole ko solve karne ke liye universal equation banana chahta hu jisko banane me mujhe problem ho rahi hai , mera ye research math se related hai sir physics se nahi isliye aap meri madad kar sakte hai plz sir .
Sir why we can not replace "a" in terms of "b" if we have "a+b=1" then "a=b-1" and then put this value in "a^2 + b^2" implies that "(b-1)^2 + b^2" and then we will have b value (1+√(3)/2 and (1-√(3)/2 and then further solve it.
Hello sir, my method was I guess similar but a lot more lengthy, for ab I did the same approach then I factored a^11+b^11 as (a+b)(a10-a9b+a8b2-a7b3+a6b4-a5b5+a4b6-a3b7+a2b8-ab9+b10) we know a+b = 1 so that goes away then I factored that entire expression in terms of ab then got a10+b10, a8+b8, a6+b6, a4+b4 and a2+b2 then accordingly, I found the respective values of the above terms, plugged in ab=-1/2 then got the final answer, is this method efficient? I would love to hear other's opinions and thoughts about this method as well :)
An easy approach a+b=1 a=1-b --------(1) a^2+b^2=2 (1-b)^2+b^2=2 ( from 1) 1-2b+2b^2=2 2b^2-2b-1=0 Using the quadratic equations formula get the value of b which comes to be (1±root(3))/2 substitute the value of b in 1 and the value of a comes to be equal with the value of b Substitute a and b in the equation a^11+b^11 and get the answer love from nepal sir
Okay but what if we just used the 2 equations to directly solve the 2 unknowns. Sub b = 1 - a into a^2 + b^2 = 2. Then the quadratic solution tells us b and a are -0.366 and 1.366. Then raise them to the power of 11 and boom same answer.
Another approach - first we find ab like in video.
Then make quadratic with a,b as roots.
Hence, x²-x-1/2=0 is our quadratic.
Now let the sequence t_n be defined as t_n = a^n + b^n
Now we know that a²= a+1/2
Multiplying a^n both sides
a^(n+2)= a^(n+1) + (1/2)* a^n
Similarly
b^(n+2)= b^(n+1) + (1/2)* b^n
Adding these we get
t_(n+2)= t_(n+1) + (1/2)* t_n
We know t_1= 1 and t_2=2.
Using the above formula we can calculate t_3,t_4 and so on uptill the required t_11.
For example
t_3= t_2+ (1/2)*t_1= 5/2
t_4= 5/2+ (1/2)*2= 7/2
t_5= 7/2+ (1/2)*(5/2) = 19/4 and so on...
Exactly I also solved like this
How this idea came in your mind
@@sankmusic9145 mind hona chahiye.
Ideas aate rhna chahiye.
Bass ...jo ki humare paas nahi hai 🗿
this was the first thing that hit my mind to solve . no need for a lengthy method used in video..
@@KoushikDas2005 toxic
Sir please take a lecture on componendo-dividendo
a+b=1
a²+b²=2
a¹¹+b¹¹=?
Now form a quadratic equation.
Clearly ab=-1/2
f(x)=x²-x-1/2
Here A=1 and B=-1/2
Now using Newton's Girand idinties for quadratic equation for symmetry polynomial.
Pᵢ=APᵢ₋₁-BPᵢ₋₂ ∀ i≥2
Pᵢ=Pᵢ₋₁+(1/2)Pᵢ₋₂
P₂=P₁+(1/2)P₀
P₂=1+(1/2)×2
P₂=2
P₃=P₂+(1/2)P₁
P₃=2+(1/2)×1
P₃=5/2= a³+b³
P₄=5/2+(1/2)×2
P₄=7/2 =a⁴+b⁴
P₅=7/2+(1/2)(5/2)
P₅=19/4=a⁵+b⁵
P₆=19/4+(1/2)(7/2)
P₆=13/2=a⁶+b⁶
P₇=13/2+(1/2)(19/4)
P₇=71/8=a⁷+b⁷
P₈=71/8+(1/2)(13/2)
P₈=97/8 =a⁸+b⁸
P₉=(97/8)+(1/2)(71/8)
P₉=265/16=a⁹+b⁹
P₁₀=265/16+(1/2)(97/8)
P₁₀=181/8=a¹⁰+b¹⁰
P₁₁=181/8+(1/2)(265/16)
P₁₁=989/32=a¹¹+b¹¹
Hence required Result=989/32
I praise your hardwork friend
I studied this method during my IMO preparation
Are you also preparing for IMO?
@Target IPhO which class, cause im currently in 9th lul
@@targetipho1301 Which board man? I never studied this.
Bro which keyboard do you use?
Longer method
a+b=1
a²+b²=2
a¹¹+b¹¹=?
Now using Newton girad identies and kth symmetric polynomial.
p₁=a+b
p₂=a²+b²
p₁₁=a¹¹+b¹¹
And
e₀=1
e₁=a+b
e₂=ab
eᵣ=0, ∀r≥3
Now
Using
keₖ=eₖ₋₁p₁-eₖ₋₂p₂+.......
e₁=e₀p₁
e₁=1
2e₂=e₁p₁-e₀p₂
2e₂=1-2
e₂=-1/2
3e₃=e₂p₁-e₁p₂+e₀p₃
0=-1/2-2+p₃
p₃=5/2
4e₄=e₃p₁-e₂p₂+e₁p₃-e₀p₄
0=1/2×2+1×5/2-p₄
p₄=7/2
Similarly
p₅=19/4
p₆=13/2
p₇=71/8
p₈=97/8
p₉=265/16
p₁₀=181/8
p₁₁=989/32
Hence
a¹¹+b¹¹=989/32
Yes I also used newton identity, it is very useful for sum of powers questions
Sir i paused video and did it with different approach. With a+b and ab i made a quadratic equation then i found a and b then a¹¹+b¹¹ became a binomial expression and as a and b were conjugate so all odd terms got cut and remaining was the answer. I got the right answer.
Can u elaborate Yr process plse
@@Beyondfutura see a+b = 1 and ab=-1/2
Using this we can for a quadratic equation , 2a²-2a-1=0 then value of a = (1+rt3)/2 and b= (1-rt3)/2
Now a¹¹ + b¹¹ = [(1+rt3)¹¹+(1-rt3)¹¹]/2¹¹ now use concepts of binomial theorem.
Calculation could me made shorter here..
Using WLOG,since from 2nd eqn we can take
a=√2 × sinӨ
b=√2 × cosӨ
Thn from first eqn find the value of Ө
How yr relating with conjugates?
I mean is binomial doable here?
Are you sayin
I ve to find 11C2,.....11C10?
You can use a different method for this using Viete's formulas. After calculating ab = -1/2, you can say that a,b are the roots of the quadratic 2x^2-2x-1=0. They are also the roots of 2x^11-2x^10-x^9=0, multiplying across by x^9, and 2x^10-2x^9-x^8=0. Adding these equations we eliminate the x^10 term to get 2x^11-3x^9-x^8=0
By repeatedly changing the degree of the leading coefficient and adding multiples to clear the second highest term, and clearing denominators, you can find that a,b are also the roots for the following polynomials:
4x^11 - 8x^8 - 3x^7 = 0
4x^11 - 11x^7 - 4x^6 = 0
8x^11 - 30x^6 - 11x^5 = 0
8x^11 - 41x^5 - 15x^4 = 0
16x^11 - 112x^4 - 41x^3 = 0
16x^11 - 153x^3 - 56x^2 = 0
32x^11 - 418x^2 - 153x = 0
Here's where the magic happens. Setting x=a,x=b, and added the expressions together, you get:
32(a^11+b^11) - 418(a^2+b^2) - 153(a+b) = 0 which gives a^11+b^11=989/32
now you made this way more harder 😢
You fucked up....
@@gamingwithpurple7880 It's actually easier only if you know thee method
Sir Taylor aur maclaurin series pe ek detailed video banaiye please... Unke proof,examples ke saath.
Proof ??😂😂😂is farzi master ka ghar jaenga usme
@@Sanatan_saarthi_1729 lagta hai apne teacher ka samman karna aata nhi hai tujhe..
Yes bro, sir proof that Taylor & maclaurin.
Mast sabal tha sir mene kar maja aagya es se meri maths ki knowledge bad jatee hai
Sir please dont stop making these type of videos where you pick an intresting question,i literally wait for them as i love math and love to be challenged by it😅
It's a terrible video. Once u got a*b. Value. U can just put it inside a+b=1 and calculate a and b. No need for all these bullshit power calculations
Tob toh school mein tumko sirf maths mein hi accha no. Aata hoga
Always love to see maths and bhannat sir together
10:10
Mujhe (a^11+b^11) generate karne ka ah...hh....... yaha pe ek ahhh...hh........... infinite years later...
Generate ho skati hai..😂😂
By the way thanks for this question i solved by another way sir 😍😍
Sir you are the best teacher in the world of MATHS 😍
🙂🙂🙂👍
Not really
@@Quagmire0810 then who is?
@jeff idk yes I agree with it but he explains math in such basic level that a student of class 9 or 10 could also understand 🥶
Arey ruko jara sabar karo....
`5a^(2)+b^(2)+3c^(2)+6ab+4bc+8ca`
Iska factorisation kaise hoga
We already know that
(A+B)^2 + (A-B)^2 = 2( A^2 + B^2)
Utilise this to arrive at
A= ( 1+✓3)/2
B= (1-✓3) /2
Then binomial, much efficient method. Yeh teacher bahut bada fraud hai bhai.😂
Sir kise pata kare ki kb squaring karni hai or kab multiply and divide both sides karna hai or kb +1-1 karna hai in any questions plz tell someone 😮
I had a different approach,
Prepare a quadratic equation using the given info. using Viete's formulae,
The expected equation is,
2x^2 - 2x - 1 = 0
since, a and b are roots,
define : S(n) = a^n + b^n
then : 2S(n) = 2S(n-1) + S(n-2)
S(n) = S(n-1) + (1/2)*S(n-2)
a^n + b^n = S(n-1) + (1/2)*S(n-2)
set n = 11,
a^11 + b^11 = S(10) + (1/2)*S(9)
This is an application of Newton's formula. After expanding the binomial, odd terms get cancelled and the answer is found to be (989/32).
Finest problem solved by Finest MATHS teacher
Sir Olympiad ke liye koi maths ki book suggest kijiye na plz 🙏🙏
Sir a aur b ke bich ke relation se unke liye solve Kiya toh ek answer a = -W aur b = -W^2 aata hai ... Where W is cube root of unity
Waha se required value -1 aati hai
-1 Nhi 1 Aati
@@Games_Era44 Ohh ha ... Thank you for correcting me
Mera a,b = (1+- √3)/2 aa rhi hai, and waha se correct answer aa rha hai 🙂🙃🙃
@@jaythakkar4298 omega hone ke liye iota ka term hona chahiye 😀
Maine a or b ki value find kar ke solve kiya. (1 + root 3 ) by 2 hua toh b (1- root 3) by 2 hoga and vice versa.
Не проще ли выразить b через a и посчитать её степени отдельно?
sir aise hi videos daal te rahiye hame bhut maze aate hain
Initially I was also going in the same direction but I thought it would long process there might be some Quick way to approach this problem but I was wrong
pehle a⁴+b⁴ nikala phir a⁸+b⁸ then a⁵+b⁵ by multiplying a²+b² and a³+b³
Then multiply a³+b³ by a⁸+b⁸ and hence a¹¹+b¹¹ is 989/32
You can easily find a-b by using 2ab and u can get values of a and b if u do power 11 its ur ans
You always honestly teach ..big fan sir 👍
Logic- increasing exponential by square until get 11
Good and highly effective steps in solving this highly complex math problem in terms of powers.. ,the question of curiosity is what is the value of (a) and (b) respectively ..... thanks...
It's not complex, someone stupid as me with basic acknowledge about algebra could easily resolve it by using a system of equations. And the value of a and b aren't just one number, they has more roots, specifically, 2 roots each one, a (sub.1)=b (sub.2)≈-0.365 and b (sub.1)=a (sub.2)≈1.365.
Insted of doing all this just solve by 2 eqn and 2 variable method
Sir aap jo question ka collection leke ate hoon woh sach mein soch ne majbur karta hein
Very interesting problem and approach sir.
Sir calculation booster video bhi bana dijiye
@gaming studio Vo video se nahi hota practice se hoga
@@Kartik-Taralekar are you Aman sir????🤔🤔🤔
@@gamingstudio829 vedic maths ki vedio toh banai hui heh sir ne
U can just find the value of a and b by using ab and a+b then substitute the values in a^11+b^11
Well nice approach
It's not possible.
@@probalroychowdhury1880 After finding a and b , solve a¹¹+b¹¹ using Binomial expansion 👍
a,b=(1+-√3)÷2
@@MA_7_22 then it will be very time consuming
@@probalroychowdhury1880use binomial theorem it won't be time consuming
The base of a number means frequency in physics. So why are there limits to frequency because they can combine only in specific patterns. Essentially time dimensions are bases. Structural coordination is f mod. Right angles mean that. Light can be modulated only in specifics of right angles. Energy is a constituent of absolute squares.
3 and 8 ka bhi combination use kar sakte hai
Good teacher . You show every step. Good.
a^+b^=1
a^2+b^2=2
a^b^=-1/2
a^k+b^k=a^k-1+b^k-1-a^b^(a^k-2+b^k-2)=a^k-1+b^k-1+1/2(a^k-2+b^k-2)
Recursive formula^:
a^^k+b^^k=a^^(k-1)+b^^(k-1) + ( a^^(k-2)+^(k-2) )/2
nextTerm = lastTerm+lastbutOne/2
a^1+b^1=1
a^2+b^2=2
a^3+b^3=2+1/2=5/2
a^4+b^4=5/2+2/2=7/2
a^5+b^5=7/2+5/4=19/4
a^6+b^6=19/4+7/4=26/4=13/2
a^7+b^7=13/2+19/8=71/8
a^8+b^8=71/8+13/4=97/8
a^9+b^9=97/8+71/16=265/16
a^10+b^10=265/16+97/16=362/16=181/8
a^11+b^11=181/8+265/32=989/32
In fact, these type of calculations are done using ana system
1Rs = 16 ana = 16'
1/2= 8 ana = 8'
1/4= 4 ana = 4'
1/8 = 2 ana = 2'
1/16 = 1 ana = 1'
1 --- 1->8'
2 --- 2->1
3 --- 2 8'->1 4'
4 --- 3 8'->1 12'
5 --- 4 12'->2 6'
6 --- 6 8'->3 4'
7 --- 8 14'->4 7'
8 --- 12 2'->6 1'
9 --- 16 9'->8 4.5'
10 --- 22 10'->11 5'
11 --- 30 14.5' === 30+29/32 === 989/32
any one understood, this ana system of calculation of recursive values
since, we need to half of last but one term,
every time, after we get a term, we half it immediately for future addition operaiton
You can first figure out the values of a and b individually then u can directly put...?
Since a+b = 1, you know that b = 1-A.
Replace b with 1-A in the second equation.
A^2 + (1-A)^2 = 2
Then solve with quadratic formula to find A.
Then you have B = 1-A.
Then calculate A^11 + B^11
Doesn't work try and see
Awesome problem! Loved it ❤
Sir kya main aapse personal me baat kar sakta hu ? Kyuki main math me relativity dipole se related ek equation banana chahta hu ye dipole thoda complicated hai aur maine ise banaya hai par is dipole ko solve karne ke liye universal equation banana chahta hu jisko banane me mujhe problem ho rahi hai , mera ye research math se related hai sir physics se nahi isliye aap meri madad kar sakte hai plz sir .
Sir Give more videos on vedic mathmatics
maza aa gaya sir. ho gaya tha solve
sir newtons theorem use karke koi bhi nth power nikalne ka recursive formulae mil jaega
I have found value of a and b then used binomial expansion to solve and get same answer in more easy way !!
Tell the whole process.
a+b=1 and a-b=root 3 find value a and b
Sir why we can not replace "a" in terms of "b" if we have "a+b=1" then "a=b-1" and then put this value in "a^2 + b^2" implies that "(b-1)^2 + b^2" and then we will have b value (1+√(3)/2 and (1-√(3)/2 and then further solve it.
Yeah now try taking a and b to the power 11,then you'll know why
Han ab usko to the power 11 le .....
Fir solve Krna band kar dega fs 😂
Sir 'A*B' ko nikal kar , a+b =1 me dalenge toh A and B ki values mil jaayegi, badme answer aa jaayega.
Itna lengthy nhi hoga
No bro a and b may be imaginary
a and b irrational aaenge . Higher powers nahi nikal paoge
Baad me 11th power kaise niklega maine bnaya mere se around 15 min me bna class 10 me hu
@@smservicesm8893 nhi imaginary nhi nikla tha (1+√3)/2 and (1-√3)/2 aya tha
@@adityamanohar2564 haa aap kaunse class me ho bhaiya,m
Sir solution can be easy if we find the value of a and b, then substitute it to a^11+b^11
I have solved in my first try . Really I enjoyed the whole steps of this problem Amazing question of algebra
Sir aap jo pen used krte h board me write krne ke liye uska naam pls btaye...usee kaha se purchase kre pls btaye sir......
Good explanation Sir
Thank you 🙏
Hello sir, my method was I guess similar but a lot more lengthy, for ab I did the same approach then I factored a^11+b^11 as
(a+b)(a10-a9b+a8b2-a7b3+a6b4-a5b5+a4b6-a3b7+a2b8-ab9+b10) we know a+b = 1 so that goes away then I factored that entire expression in terms of ab then got a10+b10, a8+b8, a6+b6, a4+b4 and a2+b2 then accordingly, I found the respective values of the above terms, plugged in ab=-1/2 then got the final answer, is this method efficient? I would love to hear other's opinions and thoughts about this method as well :)
I got the values of and b
(1+√3)/2 and (1-√3)/2..
Same
An easy approach
a+b=1
a=1-b --------(1)
a^2+b^2=2
(1-b)^2+b^2=2 ( from 1)
1-2b+2b^2=2
2b^2-2b-1=0
Using the quadratic equations formula get the value of b which comes to be (1±root(3))/2
substitute the value of b in 1 and the value of a comes to be equal with the value of b
Substitute a and b in the equation a^11+b^11 and get the answer
love from nepal sir
I have a question if you can substitute a as 1-b
Beautiful. Thank you.
Thanks for this video sir
Sir isko short me lagane ki koi trick hai ? Please bata dijiye 🙏🙏
Sir pls bring KBS. (Kaun Banega Shaharyaar) again...💕💕
Agar 1 ghante ka Olympiad hai aur 40 question hai to itna time lagana padega ek question me ?
we can also use binomial theorem
Amazing
Sir a8+b8(a3+b3) karke kar sakte hai
Sir 2023²⁰²³ ko ३५ se divide kare to remainder kya aayega video bnaye please
Sir which type of board is this and what type of marker are u using
Salute to you SIR
Great Sir
The problem needs to restrict the answer in rational in advance.
But it's really a nice blackboard.
Great man
it takes less than 2minutes to find a and b from first two equations, calculate a*11 +b*11!!!!!!!!! Why didn't I think of that?????
We can simply solve this with the help of equation a2 + b2 and a3 + b3
Where can I buy this eletronic board?
Question can also solve by binomial therom
Sir pls make a video on "all the number theory" discovered by THE GREAT RAMANUJAN
Can't we use binomial theorem?
Sir mujhe aapse request karna hai ki aap daily Olympiad ke tough sawal laya kariye please☺️☺️
Excellent
Why dont you multiply five time a2+b2 then a+b , it gives you simply a11+b11 😅
we'll have to apply binomial and thefind the values of ab multiples
Sir please one shot series shuru kijiye
Great! ❤
integral of f(x) = f(x) then f(x)=? f(x) is not e power x. please tell sir
That was so easy
love from nepal respected sir 💝
Nice sir
(a+b)^2=a^2+b^2+2ab
1=2+2ab
ab=-1/2
(a-b)=✓(2-1)=-+1
a+b+a-b=1-+1
a=0 or 1
b= 1or 0
a^11+ b^11=1
Is this method correct
make the quadratic and use newtons formula
Sir I do this question from with Newton formula
U are superb
Ingénieux ! Bravo
Sir I did this sum I am in 10th
This sum is for which standard?
9th 10th
@@msathwik8729 you are in which class?
@@shishirjha4219 10th
Thanks sir 😊
sir, i did it in another way and it's pretty much smaller.. first i found out the value of 'a' &'b' and then replaced the value in the equation
How do you find the value of a and b? Please explain
How to solve it shortly sir
Best sir
Sir agar a or b dono ka value nikale e to easily answer aa gaya.
Questions looks simple but approach is totally different,,,, i hope everyone gets understand properly😊😊😊
Okay but what if we just used the 2 equations to directly solve the 2 unknowns. Sub b = 1 - a into a^2 + b^2 = 2. Then the quadratic solution tells us b and a are -0.366 and 1.366. Then raise them to the power of 11 and boom same answer.
Yeah but in Olympiad you cannot use calculator
@@herambsharma7408fair point, so it can be left in the form you get directly from solving the quadratic
Im very curious, cause I can speak English and some sentences I understand, is it just a heavy accent or is it another language?
Both