If a and b `(ne 0)` are the roots of the quadratic `x^(2)+ax+b=0` then the least value of `x^(2)+ax+b (x in R)` is

Options

A. `-(9)/(4)`
B. `(9)/(4)`
C. `-(1)/(4)`
D. `(1)/(4)`

Answer

To find the least value of the quadratic expression \( f(x) = x^2 + ax + b \) where \( a \) and \( b \) are the roots of the quadratic equation \( x^2 + ax + b = 0 \), we can follow these steps:

### Step 1: Identify the roots and their relationships
Given that \( a \) and \( b \) are the roots of the quadratic equation \( x^2 + ax + b = 0 \), we can use Vieta's formulas:
- The sum of the roots \( a + b = -\text{coefficient of } x = -a \)
- The product of the roots \( ab = \text{constant term} = b \)

From the sum of the roots:
\[ a + b = -a \implies 2a + b = 0 \implies b = -2a \]

### Step 2: Substitute \( b \) in the quadratic expression
Now, substituting \( b \) in the expression \( f(x) \):
\[ f(x) = x^2 + ax + (-2a) = x^2 + ax - 2a \]

### Step 3: Find the vertex of the quadratic
The vertex of a quadratic function \( f(x) = Ax^2 + Bx + C \) occurs at \( x = -\frac{B}{2A} \). Here, \( A = 1 \) and \( B = a \):
\[ x = -\frac{a}{2} \]

### Step 4: Calculate the minimum value
Now, we substitute \( x = -\frac{a}{2} \) back into the function to find the minimum value:
\[ f\left(-\frac{a}{2}\right) = \left(-\frac{a}{2}\right)^2 + a\left(-\frac{a}{2}\right) - 2a \]
\[ = \frac{a^2}{4} - \frac{a^2}{2} - 2a \]
\[ = \frac{a^2}{4} - \frac{2a^2}{4} - \frac{8a}{4} \]
\[ = \frac{a^2 - 2a^2 - 8a}{4} = \frac{-a^2 - 8a}{4} \]
\[ = -\frac{a^2 + 8a}{4} \]

### Step 5: Find the minimum value in terms of \( a \)
Now, we can factor the expression \( -\frac{a^2 + 8a}{4} \):
\[ = -\frac{1}{4}(a^2 + 8a) = -\frac{1}{4}((a + 4)^2 - 16) = -\frac{(a + 4)^2}{4} + 4 \]

The minimum value occurs when \( (a + 4)^2 = 0 \), which gives \( a = -4 \). Therefore, substituting \( a = -4 \):
\[ -\frac{(0)^2}{4} + 4 = 4 \]

### Final Result
Thus, the least value of \( f(x) = x^2 + ax + b \) is:
\[ \boxed{4} \]