Let `f(x) = x2 + b_1x + c_1. g(x) = x^2 + b_2x + c_2`. Real roots of `f(x) = 0` be `alpha, beta` and real roots of `g(x) = 0` be `alpha+gamma, beta+gamma`. Least values of `f(x)` be `- 1/4`Least value of `g(x)` occurs at `x=7/2`

Options

A. `-8`
B. `-7`
C. `-6`
D. `5`

Explanation

To solve the problem, we need to analyze the given functions \( f(x) \) and \( g(x) \) and their properties.

### Step 1: Identify the functions and their properties
We have:
- \( f(x) = x^2 + b_1 x + c_1 \)
- \( g(x) = x^2 + b_2 x + c_2 \)

The real roots of \( f(x) = 0 \) are \( \alpha \) and \( \beta \), and the real roots of \( g(x) = 0 \) are \( \alpha + \gamma \) and \( \beta + \gamma \).

### Step 2: Find the least value of \( f(x) \)
The least value of a quadratic function \( ax^2 + bx + c \) occurs at \( x = -\frac{b}{2a} \). Here, \( a = 1 \), so:
\[ x = -\frac{b_1}{2} \]
The least value of \( f(x) \) is given as \( -\frac{1}{4} \). Therefore:
\[ f\left(-\frac{b_1}{2}\right) = -\frac{1}{4} \]
Calculating this:
\[ f\left(-\frac{b_1}{2}\right) = \left(-\frac{b_1}{2}\right)^2 + b_1\left(-\frac{b_1}{2}\right) + c_1 = \frac{b_1^2}{4} - \frac{b_1^2}{2} + c_1 \]
Simplifying:
\[ = \frac{b_1^2}{4} - \frac{2b_1^2}{4} + c_1 = -\frac{b_1^2}{4} + c_1 \]
Setting this equal to \( -\frac{1}{4} \):
\[ -\frac{b_1^2}{4} + c_1 = -\frac{1}{4} \]
Thus:
\[ c_1 = \frac{b_1^2}{4} - \frac{1}{4} \]

### Step 3: Find the least value of \( g(x) \)
The least value of \( g(x) \) occurs at \( x = \frac{-b_2}{2} \). We know this occurs at \( x = \frac{7}{2} \):
\[ \frac{-b_2}{2} = \frac{7}{2} \]
From this, we can find \( b_2 \):
\[ -b_2 = 7 \implies b_2 = -7 \]

### Step 4: Calculate the least value of \( g(x) \)
Now, substituting \( b_2 = -7 \) into the expression for the least value of \( g(x) \):
\[ g\left(\frac{7}{2}\right) = \left(\frac{7}{2}\right)^2 - 7\left(\frac{7}{2}\right) + c_2 \]
Calculating:
\[ = \frac{49}{4} - \frac{49}{2} + c_2 = \frac{49}{4} - \frac{98}{4} + c_2 = -\frac{49}{4} + c_2 \]
To find \( c_2 \), we need to relate it to the discriminant:
Using the relationship between \( c_1 \) and \( c_2 \), we have:
\[ c_2 = \frac{b_2^2}{4} - \frac{1}{4} = \frac{(-7)^2}{4} - \frac{1}{4} = \frac{49}{4} - \frac{1}{4} = \frac{48}{4} = 12 \]

### Step 5: Final calculation for least value of \( g(x) \)
Substituting \( c_2 \) back:
\[ g\left(\frac{7}{2}\right) = -\frac{49}{4} + 12 = -\frac{49}{4} + \frac{48}{4} = -\frac{1}{4} \]

### Conclusion
Thus, the least value of \( g(x) \) occurs at \( x = \frac{7}{2} \) and is equal to \( -\frac{1}{4} \).