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 =72`
C. ` b^(2) -b =30`
D. ` b^(2) -b =42`

Answer

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

### Step 1: Identify the roots of the quadratic equation
The roots of the quadratic equation \(x^2 + bx + 45 = 0\) can be expressed using the quadratic formula:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]
Here, \(a = 1\), \(b = b\), and \(c = 45\).

### Step 2: Express roots in terms of complex numbers
Since we are not given the value of \(b\), we will denote the roots as:
\[ x_1 = \alpha + i\beta \quad \text{and} \quad x_2 = \alpha - i\beta \]
where \(\alpha\) is the real part and \(\beta\) is the imaginary part.

### Step 3: Use Vieta's formulas
According to Vieta's formulas:
- The sum of the roots \(x_1 + x_2 = -b\)
- The product of the roots \(x_1 \cdot x_2 = 45\)

From the sum of the roots:
\[ (\alpha + i\beta) + (\alpha - i\beta) = 2\alpha = -b \quad \Rightarrow \quad \alpha = -\frac{b}{2} \]

From the product of the roots:
\[ (\alpha + i\beta)(\alpha - i\beta) = \alpha^2 + \beta^2 = 45 \]

### Step 4: Relate to the given curve
The problem states that the roots lie on the curve defined by \(|z + 1| = 2\sqrt{10}\). We can express this in terms of \(\alpha\) and \(\beta\):
\[ |(\alpha + i\beta) + 1| = 2\sqrt{10} \]
This simplifies to:
\[ |(\alpha + 1) + i\beta| = 2\sqrt{10} \]
Calculating the modulus gives:
\[ \sqrt{(\alpha + 1)^2 + \beta^2} = 2\sqrt{10} \]
Squaring both sides:
\[ (\alpha + 1)^2 + \beta^2 = 40 \]

### Step 5: Set up equations
We now have two equations:
1. \(\alpha^2 + \beta^2 = 45\) (from product of roots)
2. \((\alpha + 1)^2 + \beta^2 = 40\) (from the curve)

### Step 6: Substitute and simplify
Substituting \(\beta^2\) from the first equation into the second:
\[ (\alpha + 1)^2 + (45 - \alpha^2) = 40 \]
Expanding and simplifying:
\[ \alpha^2 + 2\alpha + 1 + 45 - \alpha^2 = 40 \]
This simplifies to:
\[ 2\alpha + 46 = 40 \quad \Rightarrow \quad 2\alpha = -6 \quad \Rightarrow \quad \alpha = -3 \]

### Step 7: Find \(b\)
We previously found that \(\alpha = -\frac{b}{2}\):
\[ -3 = -\frac{b}{2} \quad \Rightarrow \quad b = 6 \]

### Step 8: Verify options
Now we will check the options provided in the question to find which one is satisfied by \(b = 6\):
1. \(b^2 + b = 36 + 6 = 42\) (not equal to 12)
2. \(b^2 + b = 36 + 6 = 42\) (not equal to 72)
3. \(b^2 - b = 36 - 6 = 30\) (this is satisfied)
4. \(b^2 - b = 36 - 6 = 30\) (not equal to 42)

### Final Answer
The correct option is:
\[ \text{Option C: } b^2 - b = 30 \]