Consider two quadratic expressions `f(x) =ax^2+ bx + c and g (x)=ax^2+px+c,( a, b, c, p,q in R, b != p)` such that their discriminants are equal. If `f(x)= g(x)` has a root `x = alpha`, then

Options

A. `alpha` will be AM of the roots of `f(x)=0` and `g(x)=0`
B. `alpha` will be AM of the roots of `f(x)=0`
C. `alpha` will be AM of the roots of `f(x)=0` or `g(x)=0`
D. `alpha` will be AM of the roots of `g(x)=0`

Explanation

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

### Step 1: Define the Quadratic Functions
We have two quadratic functions:
- \( f(x) = ax^2 + bx + c \)
- \( g(x) = ax^2 + px + c \)

### Step 2: Calculate the Discriminants
The discriminant \( D \) of a quadratic equation \( ax^2 + bx + c \) is given by:
\[ D = b^2 - 4ac \]
For \( f(x) \):
\[ D_f = b^2 - 4ac \]
For \( g(x) \):
\[ D_g = p^2 - 4ac \]

### Step 3: Set the Discriminants Equal
Since the discriminants are equal, we have:
\[ b^2 - 4ac = p^2 - 4ac \]
Cancelling \( -4ac \) from both sides gives:
\[ b^2 = p^2 \]

### Step 4: Solve for \( b \) and \( p \)
From \( b^2 = p^2 \), we can conclude:
\[ b = p \quad \text{or} \quad b = -p \]
Since it is given that \( b \neq p \), we must have:
\[ b = -p \]

### Step 5: Set the Quadratic Functions Equal
Now, we set \( f(x) \) equal to \( g(x) \):
\[ ax^2 + bx + c = ax^2 + px + c \]
Cancelling \( ax^2 \) and \( c \) from both sides results in:
\[ bx = px \]
This simplifies to:
\[ (b - p)x = 0 \]

### Step 6: Identify the Roots
Since \( b \neq p \), we must have:
\[ x = 0 \]
This means that \( x = 0 \) is a root of the equation \( f(x) = g(x) \).

### Step 7: Find the Other Roots
Using \( b = -p \), we can rewrite the functions:
- \( f(x) = ax^2 + bx + c \)
- \( g(x) = ax^2 - bx + c \)

The roots of \( f(x) = 0 \) are:
\[ x = 0 \quad \text{and} \quad x = -\frac{b}{a} \]
The roots of \( g(x) = 0 \) are:
\[ x = 0 \quad \text{and} \quad x = \frac{p}{a} = -\frac{b}{a} \]

### Step 8: Relate the Roots to \( \alpha \)
Since \( f(x) = g(x) \) has a root \( x = \alpha \), and we have established that \( \alpha \) must be the average of the roots of \( f(x) \) and \( g(x) \):
\[ \alpha = \frac{-\frac{b}{a} + \left(-\frac{b}{a}\right)}{2} = -\frac{b}{2a} \]

### Conclusion
Thus, the root \( \alpha \) is the arithmetic mean of the roots of \( f(x) = 0 \) and \( g(x) = 0 \).