Solve $(1 - a^{2}) (x + a) - 2 a (1 - x^{2}) = 0$

x = a, \frac{- (1 + a^{2})}{2 a}

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x = 0, \frac{- (1 + a^{2})}{2 a}

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x = 0, \frac{- (1 + a^{2})}{4 a}

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x = 2 a, \frac{- (1 + a^{2})}{4 a}

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Solution

The correct option is A $x = a, \frac{- (1 + a^{2})}{2 a}$
Given, $(1 - a^{2}) (x + a) - 2 a (1 - x^{2}) = 0$
$x + a - a^{2} x - a^{3} - 2 a + 2 a x^{2} = 0$
$x - a - a^{2} x - a^{3} + 2 a x^{2} = 0$
$2 a x^{2} + x (1 - a^{2}) - (a + a^{3}) = 0$
By using formula, we have
$x = \frac{- (1 - a^{2}) \pm \sqrt{(1 - a^{2})^{2} - 4 (2 a) [- (a + a^{3})]}}{2 (2 a)}$
$= \frac{- (1 - a^{2}) \pm \sqrt{(1 + 3 a^{2})^{2}}}{4 a}$
$= \frac{- (1 - a^{2}) \pm (1 + 3 a^{2})}{4 a}$
$= \frac{- 1 + a^{2} + 1 + 3 a^{2}}{4 a}; \frac{- 1 + a^{2} - 1 - 3 a^{2}}{4 a}$
$= \frac{4 a^{2}}{4 a}; \frac{- 2 - 2 a^{2}}{4 a}$
$= a; \frac{- (1 + a^{2})}{2 a}$

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