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

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(D) Given the set $A = \{x \in R : x^2 - 5|x| + 6 = 0\}$.Since $x^2 = |x|^2$,the equation becomes $|x|^2 - 5|x| + 6 = 0$.Let $|x| = t$,then $t^2 - 5t

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$4$

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(D) Given the set $A = \{x \in R : x^2 - 5|x| + 6 = 0\}$.Since $x^2 = |x|^2$,the equation becomes $|x|^2 - 5|x| + 6 = 0$.Let $|x| = t$,then $t^2 - 5t + 6 = 0$.Factoring the quadratic equation: $(t - 2)(t - 3) = 0$.So,$|x| = 2$ or $|x| = 3$.If $|x| = 2$,then $x = 2$ or $x = -2$.If $|x| = 3$,then $x = 3$ or $x = -3$.Thus,the set $A = \{-3, -2, 2, 3\}$.The number of elements in set $A$ is $n(A) = 4$.

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

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