Key Concepts and Formulas

• Greatest Integer Function (GIF): For any real number $t$, $[t]$ denotes the greatest integer less than or equal to $t$.
• GIF Property: For any real number $x$ and integer $n$, $[x+n] = [x] + n$.
• GIF Interval: $[x] = k$, where $k$ is an integer, is equivalent to $k \le x < k+1$.

Step-by-Step Solution

Step 1: Simplify the equation using the GIF property

We are given the equation: $[x]^2 + 2[x+2] - 7 = 0$ We want to simplify the term $[x+2]$ using the property $[x+n] = [x] + n$. Here, $n=2$, which is an integer. Therefore, $[x+2] = [x] + 2$. Substituting this into the original equation, we get: $[x]^2 + 2([x] + 2) - 7 = 0$ Expanding the equation gives: $[x]^2 + 2[x] + 4 - 7 = 0$ Combining the constant terms, we get: $[x]^2 + 2[x] - 3 = 0$ This is a quadratic equation in terms of $[x]$.

Step 2: Solve the quadratic equation for [x]

Let $y = [x]$. Since $[x]$ is always an integer, $y$ must be an integer. Substituting $y$ into the equation, we have: $y^2 + 2y - 3 = 0$ We can factor this quadratic equation by finding two numbers that multiply to -3 and add to 2. These numbers are 3 and -1. Thus, we can factor the equation as: $(y + 3)(y - 1) = 0$ This equation gives us two possible solutions for $y$: $y + 3 = 0 \implies y = -3$ $y - 1 = 0 \implies y = 1$ Since $y = [x]$, the possible integer values for $[x]$ are: $[x] = 1 \quad \text{or} \quad [x] = -3$

Step 3: Determine the intervals for x

We use the property that $[x] = k$ implies $k \le x < k+1$.

• Case 1: $[x] = 1$ This means $1 \le x < 1+1$, so $1 \le x < 2$.

• Case 2: $[x] = -3$ This means $-3 \le x < -3+1$, so $-3 \le x < -2$.

The solutions for $x$ are in the intervals $[1, 2)$ and $[-3, -2)$.

Step 4: Determine if there are integral solutions

The question asks for the number of integral solutions. An integral solution is a value of $x$ that is an integer and satisfies the original equation. We examine the intervals found in Step 3 to see if they contain any integers.

• For the interval $1 \le x < 2$: The only integer in this interval is $x = 1$. Let's verify if $x=1$ is indeed a solution by substituting it back into the original equation: $[1]^2 + 2[1+2] - 7 = 1^2 + 2[3] - 7 = 1 + 2(3) - 7 = 1 + 6 - 7 = 0$ Since $0=0$, $x=1$ is an integral solution.

• For the interval $-3 \le x < -2$: The only integer in this interval is $x = -3$. Let's verify if $x=-3$ is indeed a solution by substituting it back into the original equation: $[-3]^2 + 2[-3+2] - 7 = (-3)^2 + 2[-1] - 7 = 9 + 2(-1) - 7 = 9 - 2 - 7 = 0$ Since $0=0$, $x=-3$ is an integral solution.

Therefore, there are two integral solutions: $x=1$ and $x=-3$.

Step 5: Re-evaluate the problem The problem states that the correct answer is (A) no integral solution. However, we have shown that x=1 and x=-3 are both integral solutions. There is an error in the provided "Correct Answer". We will proceed to find the answer that matches our results.

Step 6: Conclusion We found that the equation has two integral solutions, $x=1$ and $x=-3$.

Common Mistakes & Tips

• Interval Notation: Remember that $[x] = k$ means $k \le x < k+1$. Be careful to include the lower bound and exclude the upper bound.
• Verification: Always verify your solutions by plugging them back into the original equation, especially when dealing with the greatest integer function.
• Integer vs. Real: $[x]$ is always an integer, but $x$ can be any real number.

Summary

By using the properties of the greatest integer function and solving the resulting quadratic equation, we found the intervals for $x$ and then identified the integral solutions. We found two integral solutions, $x=1$ and $x=-3$. Therefore, the equation has exactly two integral solutions. The provided answer is incorrect. The correct answer should be (B).

Final Answer

The final answer is \boxed{exactly two solutions}, which corresponds to option (B).