The solution set of |x|^2-5|x|+4

Question

(C) Given the inequality: $|x|^2-5|x|+4 < 0$ Let $|x| = y$. Since $|x| \ge 0$,we have $y \ge 0$. The inequality becomes $y^2-5y+4 < 0$. Factoring the

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$(-4,-1) \cup (1,4)$

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(C) Given the inequality: $|x|^2-5|x|+4 Let $|x| = y$. Since $|x| \ge 0$,we have $y \ge 0$. The inequality becomes $y^2-5y+4 Factoring the quadratic expression: $(y-4)(y-1) This implies $1 Substituting back $y = |x|$,we get $1 This inequality is equivalent to $|x| > 1$ and $|x| For $|x| > 1$,$x \in (-\infty, -1) \cup (1, \infty)$. For $|x| Taking the intersection of these two sets,we get $x \in (-4, -1) \cup (1, 4)$.

The solution set of $|x|^2-5|x|+4 < 0$ is

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