If [x]^2-5[x]+6=0,where [x] denotes the great | vedclass

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(D) Given equation: $[x]^2-5[x]+6=0$ Let $[x] = a$. Then the equation becomes $a^2-5a+6=0$. Factoring the quadratic equation: $(a-2)(a-3)=0$. This giv

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$x \in [2,4)$

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(D) Given equation: $[x]^2-5[x]+6=0$ Let $[x] = a$. Then the equation becomes $a^2-5a+6=0$. Factoring the quadratic equation: $(a-2)(a-3)=0$. This gives $a=2$ or $a=3$. Substituting back $[x]=a$,we get $[x]=2$ or $[x]=3$. For $[x]=2$,the range of $x$ is $x \in [2,3)$. For $[x]=3$,the range of $x$ is $x \in [3,4)$. Combining these intervals,we get $x \in [2,3) \cup [3,4) = [2,4)$.

If $[x]^2-5[x]+6=0$,where $[x]$ denotes the greatest integer function,then

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