Find the value of k for which `x^(2) + k(4x + k - 1) + 2 = 0` has real and equal roots.

Answer

To find the value of \( k \) for which the equation \( x^2 + k(4x + k - 1) + 2 = 0 \) has real and equal roots, we can follow these steps:

### Step 1: Rewrite the Equation
Start by expanding the equation:
\[ x^2 + k(4x + k - 1) + 2 = 0 \]
This simplifies to:
\[ x^2 + 4kx + (k^2 - k + 2) = 0 \]

### Step 2: Identify Coefficients
From the standard form \( ax^2 + bx + c = 0 \), we identify:
- \( a = 1 \)
- \( b = 4k \)
- \( c = k^2 - k + 2 \)

### Step 3: Use the Discriminant Condition
For the quadratic equation to have real and equal roots, the discriminant \( D \) must be equal to zero:
\[ D = b^2 - 4ac = 0 \]
Substituting the values of \( a \), \( b \), and \( c \):
\[ (4k)^2 - 4(1)(k^2 - k + 2) = 0 \]

### Step 4: Simplify the Discriminant
Calculating the discriminant:
\[ 16k^2 - 4(k^2 - k + 2) = 0 \]
Distributing the \( -4 \):
\[ 16k^2 - 4k^2 + 4k - 8 = 0 \]
Combine like terms:
\[ 12k^2 + 4k - 8 = 0 \]

### Step 5: Divide the Equation
To simplify, divide the entire equation by 4:
\[ 3k^2 + k - 2 = 0 \]

### Step 6: Factor the Quadratic
Now, we need to factor \( 3k^2 + k - 2 \):
\[ 3k^2 + 3k - 2k - 2 = 0 \]
Grouping the terms:
\[ 3k(k + 1) - 2(k + 1) = 0 \]
Factoring out \( (k + 1) \):
\[ (k + 1)(3k - 2) = 0 \]

### Step 7: Solve for \( k \)
Setting each factor to zero gives:
1. \( k + 1 = 0 \) → \( k = -1 \)
2. \( 3k - 2 = 0 \) → \( k = \frac{2}{3} \)

### Conclusion
The values of \( k \) for which the equation has real and equal roots are:
\[ k = -1 \quad \text{and} \quad k = \frac{2}{3} \]

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