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Question Number 123210 by Khalmohmmad last updated on 24/Nov/20

Answered by Lordose last updated on 24/Nov/20

  (√(2+(√(2−(√(2+x)))))) = x  2+(√(2−(√(2+x)))) = x^2   2−(√(2+x)) = (x^2 −2)^2   (2−(x^2 −2)^2 )^2 = 2 + x  x^8 −8x^6 +20x^4 −16x^2 +4 − 2 − x = 0  x^8 −8x^6 +20x^4 −16x^2 −x+2=0  (x+1)(x−2)(x^6 +x^5 −5x^4 −3x^3 +7x^2 +x−1)=0  x = −1 or x = 2   ∀ {x∈Z}

2+22+x=x2+22+x=x222+x=(x22)2(2(x22)2)2=2+xx88x6+20x416x2+42x=0x88x6+20x416x2x+2=0(x+1)(x2)(x6+x55x43x3+7x2+x1)=0x=1orx=2{xZ}

Commented by MJS_new last updated on 24/Nov/20

x=−1 must be wrong because (√r)=−1 has  no solution in C  x=2 leads to (√(2+(√(2−(√(2+2))))))=(√(2+(√(2−2))))=(√2)≠2  as you should know squaring introduces  wrong solutions

x=1mustbewrongbecauser=1hasnosolutioninCx=2leadsto2+22+2=2+22=22asyoushouldknowsquaringintroduceswrongsolutions

Commented by 676597498 last updated on 24/Nov/20

true

true

Commented by MJS_new last updated on 24/Nov/20

x^6 +x^5 −5x^4 −3x^3 +7x^2 +x−1=0  (x^3 −3x+1)(x^3 +x^2 −2x−1)=0  solving both using the trigonometric method  and testing the solutions we get  x_1 =−2cos (π/9) false  x_2 =2sin (π/(18)) false  x_3 =2cos ((2π)/9) true  x_4 =−(1/3)−((2(√7))/3)cos ((π/6)+(1/3)arcsin ((√7)/(14))) false  x_5 =−(1/3)−((2(√7))/3)sin ((1/3)arcsin ((√7)/(14))) false  x_6 =−(1/3)+((2(√7))/3)sin ((π/3)+(1/3)arcsin ((√7)/(14))) false  ⇒ x=2cos ((2π)/9)

x6+x55x43x3+7x2+x1=0(x33x+1)(x3+x22x1)=0solvingbothusingthetrigonometricmethodandtestingthesolutionswegetx1=2cosπ9falsex2=2sinπ18falsex3=2cos2π9truex4=13273cos(π6+13arcsin714)falsex5=13273sin(13arcsin714)falsex6=13+273sin(π3+13arcsin714)falsex=2cos2π9

Commented by Lordose last updated on 24/Nov/20

True  Didn′t thought that

TrueDidntthoughtthat

Commented by Lordose last updated on 24/Nov/20

Thanks sir

Thankssir

Commented by Khalmohmmad last updated on 24/Nov/20

Thanks  sir

Thankssir

Answered by Dwaipayan Shikari last updated on 24/Nov/20

(√(2+(√(2−(√(2+2cosθ)))))) =2cosθ     x=2cosθ  (√(2+(√(2−2cos(θ/2))))) =2cosθ  (√(2+2sin(θ/4))) =2cosθ  (√2) (sin(θ/8)+cos(θ/8))=2cosθ  (cos((π/4)−(θ/8)))=cosθ ⇒ θ=2kπ+((2π)/9)  x=2cos((2π)/9)  As  (x is positive)

2+22+2cosθ=2cosθx=2cosθ2+22cosθ2=2cosθ2+2sinθ4=2cosθ2(sinθ8+cosθ8)=2cosθ(cos(π4θ8))=cosθθ=2kπ+2π9x=2cos2π9As(xispositive)

Commented by Khalmohmmad last updated on 24/Nov/20

Thanks  sir

Thankssir

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