Let `a, b, c` be real numbers, `a != 0.` If `alpha` is a zero of `a^2 x^2+bx+c=0, beta` is the zero of `a^2x^2-bx-c=0 and 0 < alpha < beta` then prove that the equation `a^2x^2+2bx+2c=0` has a root `gamma` that always satisfies `alpha < gamma < beta.`

Explanation

To prove that the equation \( a^2 x^2 + 2bx + 2c = 0 \) has a root \( \gamma \) that satisfies \( \alpha < \gamma < \beta \), we will follow these steps:

### Step 1: Understand the given equations and roots

We have two equations:
1. \( a^2 x^2 + bx + c = 0 \) with root \( \alpha \)
2. \( a^2 x^2 - bx - c = 0 \) with root \( \beta \)

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