For equation `[x]^2+2[x+2]-7=0, x in R` number of solution of equation is/are

Options

A. Four integer solution
B. Infinite solution
C. No solution
D. two solution

Answer

To solve the equation \([x]^2 + 2[x + 2] - 7 = 0\) where \(x \in \mathbb{R}\), we will follow these steps:

### Step 1: Rewrite the equation using the greatest integer function
The greatest integer function, denoted as \([x]\), gives the largest integer less than or equal to \(x\). We can express the equation as:

\[ [x]^2 + 2[x + 2] - 7 = 0 \]

### Step 2: Simplify the equation
Using the property of the greatest integer function, we can rewrite \([x + 2]\) as \([x] + 2\). Therefore, the equation becomes:

\[ [x]^2 + 2([x] + 2) - 7 = 0 \]

Expanding this, we get:

\[ [x]^2 + 2[x] + 4 - 7 = 0 \]

This simplifies to:

\[ [x]^2 + 2[x] - 3 = 0 \]

### Step 3: Substitute \([x]\) with a variable
Let \(t = [x]\). The equation now becomes:

\[ t^2 + 2t - 3 = 0 \]

### Step 4: Factor the quadratic equation
We can factor the quadratic equation:

\[ (t + 3)(t - 1) = 0 \]

### Step 5: Solve for \(t\)
Setting each factor equal to zero gives us:

\[ t + 3 = 0 \quad \Rightarrow \quad t = -3 \]
\[ t - 1 = 0 \quad \Rightarrow \quad t = 1 \]

### Step 6: Determine the corresponding values of \(x\)
Now we need to find the values of \(x\) corresponding to these values of \(t\):

1. For \(t = -3\):
\[ [x] = -3 \quad \Rightarrow \quad -3 \leq x < -2 \]
This gives us an infinite number of solutions in the interval \([-3, -2)\).

2. For \(t = 1\):
\[ [x] = 1 \quad \Rightarrow \quad 1 \leq x < 2 \]
This also gives us an infinite number of solutions in the interval \([1, 2)\).

### Step 7: Conclusion
Since both intervals provide infinite solutions, the total number of solutions to the equation is infinite.

Thus, the final answer is that there are infinite solutions.

---