$\frac{1 - x}{1 + x} = x$ , find $x$

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Solution

$\frac{1 - x}{1 + x} = x$

$1 - x = x + x^{2}$

$x^{2} + 2 x - 1 = 0$

We have a formula for solving quadratic equation $a x^{2} + b x + c = 0$ is

$x = \frac{- b \pm \sqrt{b^{2} - 4 a c}}{2 a}$

$= \frac{- 2 \pm \sqrt{(2)^{2} - 4 (1) (- 1)}}{2 (1)}$

$x = - 1 \pm \sqrt{2}$

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Similar questions

y = \tan^{- 1} (\frac{\sqrt{1 + x} - \sqrt{1 - x}}{\sqrt{1 + x} + \sqrt{1 - x}}), find \frac{d y}{d x}

$l o g_{e} [(1 + x)^{1 + x} . (1 - x)^{1 - x}] = a [\frac{x^{2}}{1.2} + \frac{x^{4}}{3.4} + \frac{x^{6}}{5.6} + . . .]$ . Find $a$ .

{tan}^{- 1} (\frac{1 - x}{1 + x}) = \frac{1}{2} {tan}^{- 1} x, x > 0 .

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