Let `f(x)=x^(2)+bx+c` and `g(x)=x^(2)+b_(1)x+c_(1)` Let the real roots of `f(x)=0` be `alpha, beta` and real roots of `g(x)=0` be `alpha +k, beta+k` fro same constant `k`. The least value fo `f(x)` is `-1/4` and least value of `g(x)` occurs at `x=7/2` The roots of `g(x)=0` are

Options

A. 3,4
B. `-3,4`
C. `-3,-4`
D. `3,-4`

Explanation

To solve the problem step by step, we will analyze the given functions and their properties.

### Step 1: Analyze the function \( f(x) \)
Given:
\[ f(x) = x^2 + bx + c \]
The roots of \( f(x) = 0 \) are \( \alpha \) and \( \beta \).

Using Vieta's formulas:
- The sum of the roots \( \alpha + \beta = -b \)
- The product of the roots \( \alpha \beta = c \)

We are also given that the least value of \( f(x) \) is \( -\frac{1}{4} \). The minimum value of a quadratic function \( ax^2 + bx + c \) occurs at \( x = -\frac{b}{2a} \). Here, \( a = 1 \), so:
\[ x = -\frac{b}{2} \]

Substituting this back into \( f(x) \) to find the minimum value:
\[ f\left(-\frac{b}{2}\right) = \left(-\frac{b}{2}\right)^2 + b\left(-\frac{b}{2}\right) + c = \frac{b^2}{4} - \frac{b^2}{2} + c = -\frac{b^2}{4} + c \]
Setting this equal to \( -\frac{1}{4} \):
\[ -\frac{b^2}{4} + c = -\frac{1}{4} \]
Rearranging gives:
\[ c = \frac{b^2}{4} - \frac{1}{4} \]

### Step 2: Substitute and simplify
We can express \( c \) in terms of \( b \):
\[ c = \frac{b^2 - 1}{4} \]

### Step 3: Analyze the function \( g(x) \)
Given:
\[ g(x) = x^2 + b_1 x + c_1 \]
The roots of \( g(x) = 0 \) are \( \alpha + k \) and \( \beta + k \).

Using Vieta's formulas for \( g(x) \):
- The sum of the roots \( (\alpha + k) + (\beta + k) = -b_1 \)
- The product of the roots \( (\alpha + k)(\beta + k) = c_1 \)

From the sum of the roots:
\[ \alpha + \beta + 2k = -b_1 \implies -b + 2k = -b_1 \implies b_1 = b - 2k \]

### Step 4: Minimum value of \( g(x) \)
We are given that the minimum value of \( g(x) \) occurs at \( x = \frac{7}{2} \). Thus:
\[ \frac{7}{2} = -\frac{b_1}{2} \implies b_1 = -7 \]

### Step 5: Substitute \( b_1 \) into the equation
Now substituting \( b_1 = -7 \) into the equation for \( b_1 \):
\[ -7 = b - 2k \implies b = -7 + 2k \]

### Step 6: Find \( c_1 \)
Now we need to find \( c_1 \):
\[ c_1 = (\alpha + k)(\beta + k) = \alpha \beta + k(\alpha + \beta) + k^2 = c + k(-b) + k^2 \]
Substituting \( c = \frac{b^2 - 1}{4} \):
\[ c_1 = \frac{b^2 - 1}{4} + k(b) + k^2 \]

### Step 7: Roots of \( g(x) \)
Now we can find the roots of \( g(x) = 0 \):
Using \( b_1 = -7 \):
\[ g(x) = x^2 - 7x + c_1 \]
The roots can be found using the quadratic formula:
\[ x = \frac{-(-7) \pm \sqrt{(-7)^2 - 4 \cdot 1 \cdot c_1}}{2 \cdot 1} \]
\[ x = \frac{7 \pm \sqrt{49 - 4c_1}}{2} \]

### Step 8: Conclusion
The roots of \( g(x) = 0 \) are:
\[ x = \frac{7 + \sqrt{49 - 4c_1}}{2} \quad \text{and} \quad x = \frac{7 - \sqrt{49 - 4c_1}}{2} \]