For $x^{2} - (a + 3) | x | + 4 = 0$ to have real solutions, the range of $a$ is

(- \infty, - 7] \cup [1, \infty)

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(- 3, \infty)

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(- \infty, - 7]

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[1, \infty)

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Solution

The correct option is B $[1, \infty)$
$a = \frac{x^{2} + 4}{| x |} - 3$
$= | x | + \frac{4}{| x |} - 3 = {(\sqrt{| x |} - \frac{2}{\sqrt{| x |}})}^{2} + 1$
$\Rightarrow a \geq 1$

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