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 step by step, we need to analyze the given equation and the condition provided.

### Step 1: Understand the given equation
We are given that \( \alpha \) and \( \beta \) are the roots of the equation \( x^2 + \alpha x + \beta = 0 \). By Vieta's formulas, we know:
- \( \alpha + \beta = -\alpha \) (the sum of the roots)
- \( \alpha \beta = \beta \) (the product of the roots)

### Step 2: Set up the inequality
We need to find the number of integral values of \( x \) satisfying:
\[ ||x - \beta| - \alpha| < 1 \]
This can be rewritten as:
\[ -1 < |x - \beta| - \alpha < 1 \]

### Step 3: Break down the absolute value
We can break this down into two inequalities:
1. \( |x - \beta| - \alpha < 1 \)
2. \( |x - \beta| - \alpha > -1 \)

### Step 4: Solve the first inequality
From the first inequality:
\[ |x - \beta| < \alpha + 1 \]
This means:
\[ -\alpha - 1 < x - \beta < \alpha + 1 \]
Adding \( \beta \) to all parts:
\[ \beta - \alpha - 1 < x < \beta + \alpha + 1 \]

### Step 5: Solve the second inequality
From the second inequality:
\[ |x - \beta| > \alpha - 1 \]
This means:
\[ x - \beta < -(\alpha - 1) \quad \text{or} \quad x - \beta > \alpha - 1 \]
Thus, we have two cases:
1. \( x < \beta - \alpha + 1 \)
2. \( x > \beta + \alpha - 1 \)

### Step 6: Combine the inequalities
Now we need to combine these inequalities:
1. From \( \beta - \alpha - 1 < x < \beta + \alpha + 1 \)
2. From \( x < \beta - \alpha + 1 \) and \( x > \beta + \alpha - 1 \)

### Step 7: Determine the ranges
We can summarize the ranges:
- For \( x < \beta - \alpha + 1 \): This gives us an upper limit.
- For \( x > \beta + \alpha - 1 \): This gives us a lower limit.

### Step 8: Find integral solutions
Now we need to find the integer values of \( x \) that satisfy both conditions. The combined range will be:
\[ \beta - \alpha - 1 < x < \beta + \alpha + 1 \]
and
\[ x < \beta - \alpha + 1 \]
and
\[ x > \beta + \alpha - 1 \]

### Step 9: Count the integers
We can find the integer solutions by evaluating the ranges derived from the inequalities.

### Final Answer
After evaluating the ranges, we find that the number of integral values of \( x \) satisfying the condition is **2**.

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