If `alpha and beta` are the roots of the equation `x^(2)+alpha x+beta=0` such that `alpha ne beta`, then the number of integral values of x satisfying `||x-beta|-alpha|lt1` is

Options

A. 0
B. 1
C. 2
D. more than 2

Answer

To solve the problem, we need to find the number of integral values of \( x \) that satisfy the inequality \( ||x - \beta| - \alpha| < 1 \), given that \( \alpha \) and \( \beta \) are the roots of the equation \( x^2 + \alpha x + \beta = 0 \) and \( \alpha \neq \beta \).

### Step 1: Understanding the roots
From the quadratic equation \( x^2 + \alpha x + \beta = 0 \), we know that:
- The sum of the roots \( \alpha + \beta = -\alpha \) (from Vieta's formulas).
- The product of the roots \( \alpha \beta = \beta \).

From \( \alpha + \beta = -\alpha \), we can rearrange this to get:
\[ 2\alpha + \beta = 0 \implies \beta = -2\alpha. \]

### Step 2: Substitute \( \beta \) into the inequality
Now, substituting \( \beta = -2\alpha \) into the inequality \( ||x - \beta| - \alpha| < 1 \):
\[ ||x + 2\alpha| - \alpha| < 1. \]

### Step 3: Break down the absolute value
Let \( y = |x + 2\alpha| - \alpha \). The inequality becomes:
\[ |y| < 1. \]
This implies:
\[ -1 < y < 1. \]
Thus, we have:
\[ -1 < |x + 2\alpha| - \alpha < 1. \]

### Step 4: Solve the inequalities
We can break this into two inequalities:
1. \( |x + 2\alpha| - \alpha < 1 \)
2. \( |x + 2\alpha| - \alpha > -1 \)

#### For the first inequality:
\[ |x + 2\alpha| < \alpha + 1. \]
This means:
\[ -\alpha - 1 < x + 2\alpha < \alpha + 1. \]
Rearranging gives:
\[ -\alpha - 1 - 2\alpha < x < \alpha + 1 - 2\alpha, \]
which simplifies to:
\[ -3\alpha - 1 < x < -\alpha + 1. \]

#### For the second inequality:
\[ |x + 2\alpha| > \alpha - 1. \]
This means:
\[ x + 2\alpha < -(\alpha - 1) \quad \text{or} \quad x + 2\alpha > \alpha - 1. \]
Rearranging gives:
1. \( x < -3\alpha + 1 \)
2. \( x > -\alpha - 1 \)

### Step 5: Combine the inequalities
Now we combine the results:
1. From the first inequality: \( -3\alpha - 1 < x < -\alpha + 1 \)
2. From the second inequality: \( -3\alpha + 1 > x > -\alpha - 1 \)

### Step 6: Determine the range for \( x \)
We need to find the intersection of these ranges:
1. \( -3\alpha - 1 < x < -\alpha + 1 \)
2. \( -3\alpha + 1 > x > -\alpha - 1 \)

The combined inequalities give us:
\[ -3\alpha - 1 < x < -\alpha + 1 \quad \text{and} \quad -\alpha - 1 < x < -3\alpha + 1. \]

### Step 7: Find integer solutions
To find the integer solutions, we need to evaluate the ranges:
- The lower bound is \( -3\alpha - 1 \) and the upper bound is \( -\alpha + 1 \).
- The lower bound is \( -\alpha - 1 \) and the upper bound is \( -3\alpha + 1 \).

We can find the integer values that satisfy both conditions.

### Final Result
After evaluating the ranges, we find that the integer values satisfying the inequalities are:
- \( -3 \) and \( -1 \).

Thus, the number of integral values of \( x \) satisfying the inequality is **2**.