If |x-2| geq |x-4| then x in ldots | vedclass

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(B) Given the inequality $|x-2| \geq |x-4|$.Since both sides are non-negative,we can square both sides:$(x-2)^2 \geq (x-4)^2$$x^2 - 4x + 4 \geq x^2 -

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$[3, \infty)$

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(B) Given the inequality $|x-2| \geq |x-4|$.Since both sides are non-negative,we can square both sides:$(x-2)^2 \geq (x-4)^2$$x^2 - 4x + 4 \geq x^2 - 8x + 16$Subtract $x^2$ from both sides:$-4x + 4 \geq -8x + 16$Add $8x$ to both sides:$4x + 4 \geq 16$Subtract $4$ from both sides:$4x \geq 12$Divide by $4$:$x \geq 3$Thus,$x \in [3, \infty)$.

If $|x-2| \geq |x-4|$ then $x \in \ldots$

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