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 \([x]\) denotes the greatest integer function (also known as the floor function), we can follow these steps:

### Step 1: Rewrite the equation
We start with the equation:
\[ [x]^2 + 2[x + 2] - 7 = 0 \]
We can expand \([x + 2]\) as \([x] + 2\). Thus, the equation becomes:
\[ [x]^2 + 2([x] + 2) - 7 = 0 \]
This simplifies to:
\[ [x]^2 + 2[x] + 4 - 7 = 0 \]
or
\[ [x]^2 + 2[x] - 3 = 0 \]

### Step 2: Substitute \(t = [x]\)
Let \(t = [x]\). The equation now reads:
\[ t^2 + 2t - 3 = 0 \]

### Step 3: Factor the quadratic equation
We can factor this quadratic equation:
\[ (t + 3)(t - 1) = 0 \]
Setting each factor to zero gives us:
\[ t + 3 = 0 \quad \Rightarrow \quad t = -3 \]
\[ t - 1 = 0 \quad \Rightarrow \quad t = 1 \]

### Step 4: Determine the corresponding \(x\) values
Since \(t = [x]\), we have two cases:
1. For \(t = -3\):
\[ -3 \leq x < -2 \]
2. For \(t = 1\):
\[ 1 \leq x < 2 \]

### Step 5: Identify the intervals for \(x\)
From the above cases, we find the intervals for \(x\):
- The first interval is \([-3, -2)\)
- The second interval is \([1, 2)\)

### Step 6: Count the solutions
Both intervals contain infinitely many real numbers. Thus, the total number of solutions to the equation is infinite.

### Final Answer
The number of solutions for the equation \([x]^2 + 2[x + 2] - 7 = 0\) is infinite.

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