If the quadratic expression `x^(2)-(a-1)x+(a+(1)/(4))` were to be a perfect square then a=

Options

A. 0, 6
B. 2, 4
C. 1, 8
D. 4, 8

Answer

To determine the values of \( a \) for which the quadratic expression

\[ x^2 - (a-1)x + \left(a + \frac{1}{4}\right) \]

is a perfect square, we will follow these steps:

### Step 1: Identify the form of a perfect square
A quadratic expression \( x^2 + bx + c \) is a perfect square if it can be expressed as \( (x - p)^2 \) for some \( p \). This expands to:

\[ x^2 - 2px + p^2 \]

### Step 2: Compare coefficients
From the expression \( x^2 - (a-1)x + \left(a + \frac{1}{4}\right) \), we can compare coefficients with \( x^2 - 2px + p^2 \):

- Coefficient of \( x \): \( -(a - 1) = -2p \)
- Constant term: \( a + \frac{1}{4} = p^2 \)

### Step 3: Solve for \( p \)
From the first equation, we can express \( p \):

\[ a - 1 = 2p \implies p = \frac{a - 1}{2} \]

### Step 4: Substitute \( p \) into the second equation
Substituting \( p \) into the second equation gives:

\[ a + \frac{1}{4} = \left(\frac{a - 1}{2}\right)^2 \]

### Step 5: Simplify the equation
Expanding the right side:

\[ \left(\frac{a - 1}{2}\right)^2 = \frac{(a - 1)^2}{4} = \frac{a^2 - 2a + 1}{4} \]

Setting the two sides equal:

\[ a + \frac{1}{4} = \frac{a^2 - 2a + 1}{4} \]

### Step 6: Clear the fraction
Multiply through by 4 to eliminate the fraction:

\[ 4a + 1 = a^2 - 2a + 1 \]

### Step 7: Rearrange the equation
Rearranging gives us:

\[ a^2 - 2a - 4a + 1 - 1 = 0 \implies a^2 - 6a = 0 \]

### Step 8: Factor the quadratic
Factoring out \( a \):

\[ a(a - 6) = 0 \]

### Step 9: Solve for \( a \)
Setting each factor to zero gives:

\[ a = 0 \quad \text{or} \quad a = 6 \]

### Conclusion
The values of \( a \) for which the quadratic expression is a perfect square are:

\[ \boxed{0 \text{ and } 6} \]