If the equation `x^(2)- bx+ 1 = 0` does not possess real roots, then which one of the following is correct?

Options

A. ` -3 lt b lt 3`
B. ` -2 lt b lt 2 `
C. ` b gt 2`
D. ` b lt -2`

Explanation

To determine the conditions under which the quadratic equation \( x^2 - bx + 1 = 0 \) does not possess real roots, we need to analyze the discriminant of the equation.

### Step-by-step Solution:

1. **Identify the coefficients**:
The given quadratic equation is in the form \( ax^2 + bx + c = 0 \), where:
- \( a = 1 \)
- \( b = -b \) (the coefficient of \( x \))
- \( c = 1 \)

2. **Write the discriminant**:
The discriminant \( D \) of a quadratic equation \( ax^2 + bx + c = 0 \) is given by:
\[ D = b^2 - 4ac \]
Substituting the values of \( a \), \( b \), and \( c \):
\[ D = (-b)^2 - 4 \cdot 1 \cdot 1 \]
This simplifies to:
\[ D = b^2 - 4 \]

3. **Condition for no real roots**:
For the quadratic equation to not possess real roots, the discriminant must be less than zero:
\[ D < 0 \]
Therefore, we set up the inequality:
\[ b^2 - 4 < 0 \]

4. **Solve the inequality**:
Rearranging the inequality gives:
\[ b^2 < 4 \]
Taking the square root of both sides, we find:
\[ -2 < b < 2 \]

5. **Conclusion**:
The values of \( b \) that satisfy this inequality indicate that \( b \) must be between -2 and 2 (exclusive). Thus, the correct answer is:
\[ -2 < b < 2 \]