Let [x] be the greatest integer less than or | vedclass

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(C) Given the equation $[x^2] = x + 1$. Since $[x^2]$ is an integer,$x + 1$ must be an integer,which implies $x$ is an integer. Let $x = n$,where $n \

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(C) Given the equation $[x^2] = x + 1$. Since $[x^2]$ is an integer,$x + 1$ must be an integer,which implies $x$ is an integer. Let $x = n$,where $n \in \mathbb{Z}$. The equation becomes $[n^2] = n + 1$. Since $n^2$ is an integer,$[n^2] = n^2$. So,$n^2 = n + 1$,which implies $n^2 - n - 1 = 0$. The roots of this quadratic equation are $n = \frac{1 \pm \sqrt{1 - 4(1)(-1)}}{2} = \frac{1 \pm \sqrt{5}}{2}$. Since $\frac{1 \pm \sqrt{5}}{2}$ are not integers,there is no integer $n$ that satisfies the equation. Therefore,the equation $[x^2] = x + 1$ has no solution.

Let $[x]$ be the greatest integer less than or equal to $x$ for a real number $x$. Then the equation $[x^2] = x + 1$ has:

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