If `x^(2)+bx+1=0,binR` has imaginary root z such that `|z+1|=3` then `|b|` can be:

Answer

To solve the problem, we need to find the value of \(|b|\) given that the quadratic equation \(x^2 + bx + 1 = 0\) has imaginary roots \(z\) such that \(|z + 1| = 3\).

### Step-by-Step Solution:

1. **Understanding the Roots**:
Since the roots are imaginary, we can denote them as \(z\) and its conjugate \(\overline{z}\). We can express \(z\) as \(p + iq\), where \(p\) is the real part and \(q\) is the imaginary part.

2. **Using the Condition on Roots**:
Given that \(|z + 1| = 3\), we can write:
\[ |(p + iq) + 1| = 3 \]
This simplifies to:
\[ |(p + 1) + iq| = 3 \]

3. **Finding the Modulus**:
The modulus can be calculated as:
\[ \sqrt{(p + 1)^2 + q^2} = 3 \]
Squaring both sides gives:
\[ (p + 1)^2 + q^2 = 9 \]

4. **Using the Product of Roots**:
From Vieta's formulas, we know:
- The sum of the roots \(z + \overline{z} = -b\)
- The product of the roots \(z \cdot \overline{z} = 1\)

The product gives us:
\[ p^2 + q^2 = 1 \]

5. **Substituting for \(q^2\)**:
We can express \(q^2\) from the product of roots:
\[ q^2 = 1 - p^2 \]

6. **Substituting in the Modulus Equation**:
Substitute \(q^2\) into the modulus equation:
\[ (p + 1)^2 + (1 - p^2) = 9 \]
Expanding this gives:
\[ p^2 + 2p + 1 + 1 - p^2 = 9 \]
Simplifying leads to:
\[ 2p + 2 = 9 \]
Thus:
\[ 2p = 7 \quad \Rightarrow \quad p = \frac{7}{2} \]

7. **Finding \(|b|\)**:
We know that:
\[ |b| = |-(z + \overline{z})| = |-(2p)| = 2|p| = 2 \times \frac{7}{2} = 7 \]

### Final Result:
Thus, \(|b| = 7\).