Roots of the equation `x^2 + bx + 45 = 0`, `b in R` lie on the curve `|z + 1| = 2sqrt(10) , where z is a complex number then

Options

A. `b^2 + b = 12`
B. `b^2 - b = 30`
C. `b^2 - b = 36`
D. `b^2 + b = 30`

Explanation

To solve the problem, we need to analyze the roots of the quadratic equation \(x^2 + bx + 45 = 0\) and their relationship with the given curve defined by \(|z + 1| = 2\sqrt{10}\).

### Step-by-Step Solution:

1. **Identify the Roots:**
The roots of the equation \(x^2 + bx + 45 = 0\) can be denoted as \(\alpha + i\beta\) and \(\alpha - i\beta\), where \(\alpha\) is the real part and \(\beta\) is the imaginary part.

2. **Sum of the Roots:**
According to Vieta's formulas, the sum of the roots is given by:
\[ \alpha + i\beta + \alpha - i\beta = -\frac{b}{1} = -b \]
This simplifies to:
\[ 2\alpha = -b \quad \Rightarrow \quad \alpha = -\frac{b}{2} \]

3. **Magnitude Condition:**
The condition given is \(|z + 1| = 2\sqrt{10}\). Substituting \(z = \alpha + i\beta\), we have:
\[ |(\alpha + i\beta) + 1| = 2\sqrt{10} \]
This can be rewritten as:
\[ |(\alpha + 1) + i\beta| = 2\sqrt{10} \]
The magnitude can be expressed as:
\[ \sqrt{(\alpha + 1)^2 + \beta^2} = 2\sqrt{10} \]

4. **Squaring Both Sides:**
Squaring both sides gives:
\[ (\alpha + 1)^2 + \beta^2 = 40 \]

5. **Product of the Roots:**
The product of the roots is given by:
\[ (\alpha + i\beta)(\alpha - i\beta) = \alpha^2 + \beta^2 = \frac{45}{1} = 45 \]

6. **Setting Up Equations:**
Now we have two equations:
- From the magnitude condition:
\[ (\alpha + 1)^2 + \beta^2 = 40 \quad \text{(1)} \]
- From the product of the roots:
\[ \alpha^2 + \beta^2 = 45 \quad \text{(2)} \]

7. **Subtracting Equations:**
Subtract equation (2) from equation (1):
\[ (\alpha + 1)^2 + \beta^2 - (\alpha^2 + \beta^2) = 40 - 45 \]
This simplifies to:
\[ (\alpha + 1)^2 - \alpha^2 = -5 \]
Expanding gives:
\[ \alpha^2 + 2\alpha + 1 - \alpha^2 = -5 \]
Thus:
\[ 2\alpha + 1 = -5 \quad \Rightarrow \quad 2\alpha = -6 \quad \Rightarrow \quad \alpha = -3 \]

8. **Finding \(b\):**
Since \(\alpha = -\frac{b}{2}\), we have:
\[ -3 = -\frac{b}{2} \quad \Rightarrow \quad b = 6 \]

9. **Verifying Options:**
We need to check which option satisfies \(b\):
- \(b^2 + b = 36 + 6 = 42\) (not an option)
- \(b^2 - b = 36 - 6 = 30\) (this is an option)
- \(b^2 - 36 = 0\) (not an option)
- \(b^2 + b - 30 = 0\) (not an option)

The correct option is \(b^2 - b = 30\).

### Final Answer:
The value of \(b\) is \(6\), and the correct option is \(b^2 - b = 30\).