The values of `k` for which the expression `kx^2 + (k + 1)x + 2` will be a perfect square of linear factor are (i) `3+-2sqrt2` (ii)`4+-2sqrt2` (iii)`6` (iv) `5`

Options

A. `3 + 2sqrt2`
B. `3 - 3sqrt2`
C. `3 + 3sqrt2`
D. `3 - 2sqrt2`

Explanation

To determine the values of \( k \) for which the expression \( kx^2 + (k + 1)x + 2 \) is a perfect square of a linear factor, we need to follow these steps:

### Step 1: Understand the condition for a perfect square
A quadratic expression \( ax^2 + bx + c \) is a perfect square if its discriminant \( D \) is equal to zero. The discriminant is given by the formula:
\[ D = b^2 - 4ac \]

### Step 2: Identify coefficients
For the expression \( kx^2 + (k + 1)x + 2 \):
- \( a = k \)
- \( b = k + 1 \)
- \( c = 2 \)

### Step 3: Calculate the discriminant
Substituting the coefficients into the discriminant formula:
\[ D = (k + 1)^2 - 4(k)(2) \]
\[ D = (k + 1)^2 - 8k \]

### Step 4: Set the discriminant to zero
To find the values of \( k \) for which the expression is a perfect square, set the discriminant equal to zero:
\[ (k + 1)^2 - 8k = 0 \]

### Step 5: Expand and simplify
Expanding the left-hand side:
\[ k^2 + 2k + 1 - 8k = 0 \]
\[ k^2 - 6k + 1 = 0 \]

### Step 6: Solve the quadratic equation
Now we will use the quadratic formula to solve for \( k \):
\[ k = \frac{-b \pm \sqrt{D}}{2a} \]
Here, \( a = 1 \), \( b = -6 \), and \( D = 1^2 - 4 \cdot 1 \cdot 1 = 36 - 4 = 32 \).

Calculating the roots:
\[ k = \frac{6 \pm \sqrt{32}}{2} \]
\[ k = \frac{6 \pm 4\sqrt{2}}{2} \]
\[ k = 3 \pm 2\sqrt{2} \]

### Step 7: List the values of \( k \)
Thus, the values of \( k \) for which the expression is a perfect square are:
\[ k = 3 + 2\sqrt{2} \quad \text{and} \quad k = 3 - 2\sqrt{2} \]

### Conclusion
The correct option is (i) \( 3 \pm 2\sqrt{2} \).

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