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(D) Given the equation: $[x]^{2} + 2[x + 2] - 7 = 0$.Using the property $[x + n] = [x] + n$ for any integer $n$,we have $[x + 2] = [x] + 2$.Substituti

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infinitely many solutions

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(D) Given the equation: $[x]^{2} + 2[x + 2] - 7 = 0$.Using the property $[x + n] = [x] + n$ for any integer $n$,we have $[x + 2] = [x] + 2$.Substituting this into the equation:$[x]^{2} + 2([x] + 2) - 7 = 0$$[x]^{2} + 2[x] + 4 - 7 = 0$$[x]^{2} + 2[x] - 3 = 0$Let $y = [x]$. Then $y^{2} + 2y - 3 = 0$.$(y + 3)(y - 1) = 0$.So,$[x] = 1$ or $[x] = -3$.If $[x] = 1$,then $x \in [1, 2)$.If $[x] = -3$,then $x \in [-3, -2)$.Since the solution set is $[1, 2) \cup [-3, -2)$,there are infinitely many real values of $x$ that satisfy the equation.

Let $[t]$ denote the greatest integer $\leq t$. Then the equation in $x$,$[x]^{2} + 2[x + 2] - 7 = 0$,has

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