let-n-1-and-2n-2-2n-1-solve-for-real-numbers-x-2-x-2-x-2-n-

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

let n ⩾ 1 and λ = {2 n}^{2} - 2 n + 1

solve for real numbers :

\sqrt{λ + x^{2}} - \sqrt{λ - x^{2}} = \frac{x^{2}}{n}

Commented by MathSh last updated on 13/Oct/21

Thank you dear S er

Solution if possible please

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

x^{2} ⩽ λ = 2 n (n - 1) + 1

2 λ - 2 \sqrt{λ^{2} - x^{4}} = \frac{x^{4}}{n^{2}}

λ - \frac{x^{4}}{2 n^{2}} = \sqrt{λ^{2} - x^{4}}

λ^{2} + \frac{x^{8}}{4 n^{4}} - \frac{λ}{n^{2}} x^{4} = λ^{2} - x^{4}

x^{8} = 4 n^{2} (λ - n^{2}) x^{4}

\Rightarrow x = 0

o r

\Rightarrow x^{4} = 4 n^{2} (λ - n^{2}) = 4 n^{2} {(n - 1)}^{2}

\Rightarrow x^{2} = 2 n (n - 1) = λ - 1 < λ ✓

\Rightarrow x = \pm \sqrt{2 n (n - 1)}

Commented by MathSh last updated on 13/Oct/21

Perfect dear S er, thankyou

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