Let `a,b,c in R and a ne 0`. If `alpha` is a root `a^(2) x^(2) +bx+c=0, beta` is a root of `a^(2) x^(2)-bx-c=0 and 0 lt alpha lt beta`, then the equation `a^(2)x^(2)+2bx+2c=0` has a root `gamma` that always satisfies

Options

A. `gamma = alpha`
B. `gamma = beta`
C. `gamma =(alpha +beta)//2`
D. `alpha lt gamma lt beta`

Answer

To solve the problem step-by-step, we will analyze the given quadratic equations and their roots.

### Step 1: Understand the Given Equations
We have two quadratic equations:
1. \( a^2 x^2 + bx + c = 0 \) with root \( \alpha \)
2. \( a^2 x^2 - bx - c = 0 \) with root \( \beta \)

### Step 2: Write the Conditions for Roots
Since \( \alpha \) is a root of the first equation, we can write:
\[ a^2 \alpha^2 + b \alpha + c = 0 \tag{1} \]
Similarly, since \( \beta \) is a root of the second equation, we can write:
\[ a^2 \beta^2 - b \beta - c = 0 \tag{2} \]

### Step 3: Analyze the New Equation
We need to analyze the equation:
\[ a^2 x^2 + 2bx + 2c = 0 \]
We will denote this function as \( f(x) = a^2 x^2 + 2bx + 2c \).

### Step 4: Evaluate \( f(\alpha) \)
Substituting \( \alpha \) into \( f(x) \):
\[ f(\alpha) = a^2 \alpha^2 + 2b \alpha + 2c \]
From equation (1), we know that:
\[ a^2 \alpha^2 + b \alpha + c = 0 \]
Thus, we can express \( f(\alpha) \) as:
\[ f(\alpha) = (a^2 \alpha^2 + b \alpha + c) + b \alpha + c = 0 + b \alpha + c = b \alpha + c \]

### Step 5: Evaluate \( f(\beta) \)
Now substituting \( \beta \) into \( f(x) \):
\[ f(\beta) = a^2 \beta^2 + 2b \beta + 2c \]
From equation (2), we have:
\[ a^2 \beta^2 - b \beta - c = 0 \]
Thus, we can express \( f(\beta) \) as:
\[ f(\beta) = (a^2 \beta^2 - b \beta - c) + 3b \beta + 3c = 0 + 3b \beta + 3c = 3b \beta + 3c \]

### Step 6: Analyze the Product \( f(\alpha) \cdot f(\beta) \)
Now, we can analyze the product:
\[ f(\alpha) \cdot f(\beta) = (b \alpha + c)(3b \beta + 3c) \]
This product will help us determine the nature of the roots of the equation \( a^2 x^2 + 2bx + 2c = 0 \).

### Step 7: Determine the Nature of the Root \( \gamma \)
Since \( 0 < \alpha < \beta \), we can conclude that:
1. \( f(\alpha) < 0 \) (as \( \alpha \) is less than the root)
2. \( f(\beta) > 0 \) (as \( \beta \) is greater than the root)

Thus, by the Intermediate Value Theorem, there exists a root \( \gamma \) such that:
\[ \alpha < \gamma < \beta \]

### Conclusion
The root \( \gamma \) of the equation \( a^2 x^2 + 2bx + 2c = 0 \) always satisfies:
\[ \alpha < \gamma < \beta \]