If A = {x in mathbb{R} : x^2 + 5|x| + 6 = 0}, | vedclass

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(A) We are given the set $A = \{x \in \mathbb{R} : x^2 + 5|x| + 6 = 0\}$.Since $x^2 = |x|^2$,the equation can be written as $|x|^2 + 5|x| + 6 = 0$.Let

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(A) We are given the set $A = \{x \in \mathbb{R} : x^2 + 5|x| + 6 = 0\}$.Since $x^2 = |x|^2$,the equation can be written as $|x|^2 + 5|x| + 6 = 0$.Let $|x| = t$,where $t \ge 0$. The equation becomes $t^2 + 5t + 6 = 0$.Factoring the quadratic,we get $(t + 2)(t + 3) = 0$.This gives $t = -2$ or $t = -3$.Since $t = |x|$ must be non-negative $(t \ge 0)$,there are no real values of $x$ that satisfy these conditions.Therefore,the set $A$ is an empty set,i.e.,$A = \emptyset$.Thus,the number of elements in $A$ is $n(A) = 0$.

If $A = \{x \in \mathbb{R} : x^2 + 5|x| + 6 = 0\}$,then $n(A) = $

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