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Question Number 156575 by MathSh last updated on 12/Oct/21

let  n≥1  and  λ=2n^2 -2n+1  solve for real numbers:  (√(λ + x^2 )) - (√(λ - x^2 )) = (x^2 /n)

$$\mathrm{let}\:\:\mathrm{n}\geqslant\mathrm{1}\:\:\mathrm{and}\:\:\lambda=\mathrm{2n}^{\mathrm{2}} -\mathrm{2n}+\mathrm{1} \\ $$$$\mathrm{solve}\:\mathrm{for}\:\mathrm{real}\:\mathrm{numbers}: \\ $$$$\sqrt{\lambda\:+\:\mathrm{x}^{\mathrm{2}} }\:-\:\sqrt{\lambda\:-\:\mathrm{x}^{\mathrm{2}} }\:=\:\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{n}} \\ $$

Commented by MathSh last updated on 13/Oct/21

Thank you dear Ser  Solution if possible please

$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{dear}\:\boldsymbol{\mathrm{S}}\mathrm{er} \\ $$$$\mathrm{Solution}\:\mathrm{if}\:\mathrm{possible}\:\mathrm{please} \\ $$

Answered by mr W last updated on 13/Oct/21

x^2 ≤λ=2n(n−1)+1  2λ−2(√(λ^2 −x^4 ))=(x^4 /n^2 )  λ−(x^4 /(2n^2 ))=(√(λ^2 −x^4 ))  λ^2 +(x^8 /(4n^4 ))−(λ/n^2 )x^4 =λ^2 −x^4   x^8 =4n^2 (λ−n^2 )x^4   ⇒x=0  or  ⇒x^4 =4n^2 (λ−n^2 )=4n^2 (n−1)^2   ⇒x^2 =2n(n−1)=λ−1<λ ✓  ⇒x=±(√(2n(n−1)))

$${x}^{\mathrm{2}} \leqslant\lambda=\mathrm{2}{n}\left({n}−\mathrm{1}\right)+\mathrm{1} \\ $$$$\mathrm{2}\lambda−\mathrm{2}\sqrt{\lambda^{\mathrm{2}} −{x}^{\mathrm{4}} }=\frac{{x}^{\mathrm{4}} }{{n}^{\mathrm{2}} } \\ $$$$\lambda−\frac{{x}^{\mathrm{4}} }{\mathrm{2}{n}^{\mathrm{2}} }=\sqrt{\lambda^{\mathrm{2}} −{x}^{\mathrm{4}} } \\ $$$$\lambda^{\mathrm{2}} +\frac{{x}^{\mathrm{8}} }{\mathrm{4}{n}^{\mathrm{4}} }−\frac{\lambda}{{n}^{\mathrm{2}} }{x}^{\mathrm{4}} =\lambda^{\mathrm{2}} −{x}^{\mathrm{4}} \\ $$$${x}^{\mathrm{8}} =\mathrm{4}{n}^{\mathrm{2}} \left(\lambda−{n}^{\mathrm{2}} \right){x}^{\mathrm{4}} \\ $$$$\Rightarrow{x}=\mathrm{0} \\ $$$${or} \\ $$$$\Rightarrow{x}^{\mathrm{4}} =\mathrm{4}{n}^{\mathrm{2}} \left(\lambda−{n}^{\mathrm{2}} \right)=\mathrm{4}{n}^{\mathrm{2}} \left({n}−\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\Rightarrow{x}^{\mathrm{2}} =\mathrm{2}{n}\left({n}−\mathrm{1}\right)=\lambda−\mathrm{1}<\lambda\:\checkmark \\ $$$$\Rightarrow{x}=\pm\sqrt{\mathrm{2}{n}\left({n}−\mathrm{1}\right)} \\ $$

Commented by MathSh last updated on 13/Oct/21

Perfect dear Ser, thankyou

$$\mathrm{Perfect}\:\mathrm{dear}\:\boldsymbol{\mathrm{S}}\mathrm{er},\:\mathrm{thankyou} \\ $$

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