How many distinct real solutions does the equation $((x^{2} - 2)^{2} - 5)^{2} = 1$ have ?

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Solution

The correct option is D 5
This equation is equivalent to
$(x^{2} - 2)^{2} - 5 = 1$ or $(x^{2} - 2)^{2} - 5 = - 1$

The first is equivalent to $x^{2} - 2 = \sqrt{6}$ or $x^{2} - 2 = - \sqrt{6}$ ,
$x = \pm (2 + \sqrt{6})$ or $x^{2} \neq - \sqrt{6} + 2$
with $2$ and $0$ solutions respectively (since - $\sqrt{6}$ + 2 < 0).

The latter is equivalent to $x^{2} - 2 = 2$ or $x^{2} - 2 = - 2,$
$x = \pm 2$ or $x = 0$
with $2$ and $1$ solution(s) respectively.

So we have $2 + 0 + 2 + 1 = 5$ solutions in total.

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