If A = {x : x^2 - 5x + 6 = 0}, B = {2, 4}, C | vedclass

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(A) Given $A = \{x : x^2 - 5x + 6 = 0\}$.Solving the quadratic equation $x^2 - 5x + 6 = 0$,we get $(x - 2)(x - 3) = 0$,so $A = \{2, 3\}$.Given $B = \{

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$\{(2, 4), (3, 4)\}$

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(A) Given $A = \{x : x^2 - 5x + 6 = 0\}$.Solving the quadratic equation $x^2 - 5x + 6 = 0$,we get $(x - 2)(x - 3) = 0$,so $A = \{2, 3\}$.Given $B = \{2, 4\}$ and $C = \{4, 5\}$.The intersection $B \cap C$ is the set of common elements,so $B \cap C = \{4\}$.Now,the Cartesian product $A 	imes (B \cap C) = \{2, 3\} 	imes \{4\}$.This results in the set of ordered pairs $\{(2, 4), (3, 4)\}$.

If $A = \{x : x^2 - 5x + 6 = 0\}, B = \{2, 4\}, C = \{4, 5\}$,then find $A \times (B \cap C)$.

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