If (x + 2)(x + 3b) = c has roots `alpha,beta` , then the roots of `(x + alpha)(x+beta) + c = 0` are

Options

A. a,b
B. -a, -b
C. -2, -3b
D. 2a, 3b

Answer

To solve the problem step by step, we start with the given equation and analyze it to find the roots of the new equation.

### Step 1: Write the given equation
The given equation is:
\[ (x + 2)(x + 3b) = c \]

### Step 2: Expand the equation
Expanding the left-hand side, we get:
\[ x^2 + 3bx + 2x + 6b = c \]
This simplifies to:
\[ x^2 + (3b + 2)x + (6b - c) = 0 \]

### Step 3: Identify the roots
Let the roots of the equation be \(\alpha\) and \(\beta\). According to Vieta's formulas:
- The sum of the roots \(\alpha + \beta = -(3b + 2)\)
- The product of the roots \(\alpha \cdot \beta = 6b - c\)

### Step 4: Write the new equation
The new equation we need to solve is:
\[ (x + \alpha)(x + \beta) + c = 0 \]
Expanding this, we have:
\[ x^2 + (\alpha + \beta)x + \alpha \beta + c = 0 \]

### Step 5: Substitute the values of \(\alpha + \beta\) and \(\alpha \beta\)
Substituting the values we found:
\[ x^2 + (-(3b + 2))x + (6b - c + c) = 0 \]
This simplifies to:
\[ x^2 - (3b + 2)x + 6b = 0 \]

### Step 6: Identify the coefficients
Now we can identify the coefficients:
- Coefficient of \(x\) (sum of roots) = \(-(3b + 2)\)
- Constant term (product of roots) = \(6b\)

### Step 7: Find the roots
To find the roots, we can use the quadratic formula:
\[ x = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A} \]
where \(A = 1\), \(B = -(3b + 2)\), and \(C = 6b\).

Calculating the roots:
\[ x = \frac{3b + 2 \pm \sqrt{(3b + 2)^2 - 4 \cdot 1 \cdot 6b}}{2 \cdot 1} \]
Calculating the discriminant:
\[ (3b + 2)^2 - 24b = 9b^2 + 12b + 4 - 24b = 9b^2 - 12b + 4 \]
Thus, the roots are:
\[ x = \frac{3b + 2 \pm \sqrt{9b^2 - 12b + 4}}{2} \]

### Step 8: Determine the roots
The roots of the equation \(x^2 - (3b + 2)x + 6b = 0\) can be expressed as:
\[ x = \frac{3b + 2 \pm \sqrt{(3b - 2)^2}}{2} \]
This gives us the roots:
\[ x = 3b \quad \text{and} \quad x = 2 \]

### Conclusion
Thus, the roots of the equation \((x + \alpha)(x + \beta) + c = 0\) are \(2\) and \(3b\).

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