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Question Number 194637 by manxsol last updated on 12/Jul/23

  x+y=1  x^2 +y^2 =2  x^(11) +y^(11) =?

x+y=1x2+y2=2x11+y11=?

Commented by Rasheed.Sindhi last updated on 14/Jul/23

See Q#191527

You can't use 'macro parameter character #' in math mode

Answered by Rasheed.Sindhi last updated on 12/Jul/23

y=1−x⇒y^2 =(1−x)^2   y^2 =2−x^2   1−2x+x^2 =2−x^2   2x^2 −2x−1=0  x=((2±(√(4−8)))/4)=((2±2i)/4)=((1±i)/2)  y=1−x=1−((1±i)/2)=((2−1∓i)/2)=((1∓i)/2)  x+y=((1±i)/2)+((1∓i)/2)=(2/2)=1   satisfy x+y=1  x^2 +y^2 =(((1±i)/2))^2 +(((1∓i)/2))^2                =((2i)/4)+((−2i)/4)=0  Do not satisfy x^2 +y^2 =2  x^(11) +y^(11) =(((1±i)/2))^(11) +(((1∓i)/2))^(11)                    =((−1)/(32))       ???

y=1xy2=(1x)2y2=2x212x+x2=2x22x22x1=0x=2±484=2±2i4=1±i2y=1x=11±i2=21i2=1i2x+y=1±i2+1i2=22=1satisfyx+y=1x2+y2=(1±i2)2+(1i2)2=2i4+2i4=0Donotsatisfyx2+y2=2x11+y11=(1±i2)11+(1i2)11=132???

Commented by Tinku Tara last updated on 13/Jul/23

2x^2 −2x−1=0  ⇒x=((2±(√(4+8)))/4)=((1±(√3))/2)  y=1−x=((1∓(√3))/2)  Mistake in your solution −8 should  be +8  (((1+(√3))/2))^(11) +(((1−(√3))/2))^(11)

2x22x1=0x=2±4+84=1±32y=1x=132Mistakeinyoursolution8shouldbe+8(1+32)11+(132)11

Commented by Rasheed.Sindhi last updated on 13/Jul/23

Thanks sir for pointing out my  mistake!

Thankssirforpointingoutmymistake!

Answered by Rasheed.Sindhi last updated on 12/Jul/23

x+y=1, x^2 +y^2 =2 , x^(11) +y^(11) =?  (x+y)^2 =(1)^2   x^2 +y^2 +2xy=1  2+2xy=1  xy=−(1/2)  (x+y)(x^2 +y^2 )=(1)(2)  x^3 +y^3 +xy(x+y)=2  x^3 +y^3 +(−(1/2))(1)=2  x^3 +y^3 =2+(1/2)=(5/2)  (x^3 +y^3 )(x+y)=((5/2))(1)  x^4 +y^4 +xy(x^2 +y^2 )=(5/2)  x^4 +y^4 +(−(1/2))(2)=(5/2)  x^4 +y^4 =(7/2)  (x^3 +y^3 )(x^2 +y^2 )=((5/2))(2)  x^5 +y^5 +x^2 y^2 (x+y)=5  x^5 +y^5 +(−(1/2))^2 (1)=5  x^5 +y^5 =5−(1/4)=((19)/4)  (x^4 +y^4 )^2 =((7/2))^2   x^8 +y^8 +2x^4 y^4 =((49)/4)  x^8 +y^8 +2(−(1/2))^4 =((49)/4)  x^8 +y^8 =((49)/4)−(1/8)=((97)/8)  (x^8 +y^8 )(x^3 +y^3 )=(((97)/8))((5/2))  x^(11) +y^(11) +x^3 y^3 (x^5 +y^5 )=((485)/(16))  x^(11) +y^(11) +(−(1/2))^3 (((19)/4))=((485)/(16))  x^(11) +y^(11) =((485)/(16))+((19)/(32))=((970+19)/(32))=((989)/(32))

x+y=1,x2+y2=2,x11+y11=?(x+y)2=(1)2x2+y2+2xy=12+2xy=1xy=12(x+y)(x2+y2)=(1)(2)x3+y3+xy(x+y)=2x3+y3+(12)(1)=2x3+y3=2+12=52(x3+y3)(x+y)=(52)(1)x4+y4+xy(x2+y2)=52x4+y4+(12)(2)=52x4+y4=72(x3+y3)(x2+y2)=(52)(2)x5+y5+x2y2(x+y)=5x5+y5+(12)2(1)=5x5+y5=514=194(x4+y4)2=(72)2x8+y8+2x4y4=494x8+y8+2(12)4=494x8+y8=49418=978(x8+y8)(x3+y3)=(978)(52)x11+y11+x3y3(x5+y5)=48516x11+y11+(12)3(194)=48516x11+y11=48516+1932=970+1932=98932

Answered by deleteduser1 last updated on 13/Jul/23

(x+y)^2 −2xy=2⇒xy=((−1)/2)  x^(11) +y^(11) =(x^6 +y^6 )(x^5 +y^5 )−(xy)^5 (x+y)  (x^6 +y^6 )=[x^2 +y^2 ][(x^2 +y^2 )^2 −3(xy)^2 ]=2(4−(3/4))=((26)/4)  x^5 +y^5 =(x^2 +y^2 )[(x+y){(x+y)^2 −3xy}]−(xy)^2 (x+y)  =2[1+(3/2)]−(1/4)=((19)/4)  ⇒x^(11) +y^(11) =((26)/4)×((19)/4)+(1/(32))=((989)/(32))

(x+y)22xy=2xy=12x11+y11=(x6+y6)(x5+y5)(xy)5(x+y)(x6+y6)=[x2+y2][(x2+y2)23(xy)2]=2(434)=264x5+y5=(x2+y2)[(x+y){(x+y)23xy}](xy)2(x+y)=2[1+32]14=194x11+y11=264×194+132=98932

Answered by manxsol last updated on 15/Jul/23

thanks Rasheed,AST,Tinku tara  my solution  r(cosθ+sinθ)=1  r=(√2)  sin(θ+45)=sin(150)  θ=105  (cos105^(11) +sin105^(11) )  (√2)^(11) [((((√6)+(√2))/4))^(11) −((((√6)−(√2))/4))^(11) ]  (√2)^(11) .(((√2)^(11) )/2^(22) )[((√3)+1)^(11) −((√3) −1)^(11) ]  (1/(2048))[2Σ_1 ^6 C_(2k−1) ^(11) 3^(6−k) ]  (3^6 /(1024))Σ_1 ^6 C_(2k−1) ^(11)  3^(−k)   30.90625=(((1+(√3))/2))^(11) +(((1−(√3))/2))^(11)

thanksRasheed,AST,Tinkutaramysolutionr(cosθ+sinθ)=1r=2sin(θ+45)=sin(150)θ=105(cos10511+sin10511)211[(6+24)11(624)11]211.211222[(3+1)11(31)11]12048[261C2k11136k]36102461C2k1113k30.90625=(1+32)11+(132)11

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