Roots of the quadratic equation `x^(2) + x- (a+1) (a+2)= 0` are ____

Options

A. `-(a+1), (a+2)`
B. `(a+1), -(a+2)`
C. `(a+1), (a+2)`
D. `-(a+1), -(a+2)`

Answer

To find the roots of the quadratic equation \( x^2 + x - (a+1)(a+2) = 0 \), we will follow these steps:

### Step 1: Identify coefficients
The given equation is in the standard form of a quadratic equation \( ax^2 + bx + c = 0 \). Here, we can identify:
- \( a = 1 \)
- \( b = 1 \)
- \( c = -(a+1)(a+2) \)

### Step 2: Write the quadratic formula
The quadratic formula to find the roots of the equation is given by:
\[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \]

### Step 3: Substitute the values of \( a \), \( b \), and \( c \)
Substituting the identified values into the quadratic formula:
\[ x = \frac{-1 \pm \sqrt{1^2 - 4 \cdot 1 \cdot (-(a+1)(a+2))}}{2 \cdot 1} \]

### Step 4: Simplify the expression under the square root
Calculate \( b^2 - 4ac \):
\[ b^2 - 4ac = 1 + 4(a+1)(a+2) \]
Now, expand \( 4(a+1)(a+2) \):
\[ 4(a+1)(a+2) = 4(a^2 + 3a + 2) = 4a^2 + 12a + 8 \]
Thus,
\[ b^2 - 4ac = 1 + 4a^2 + 12a + 8 = 4a^2 + 12a + 9 \]

### Step 5: Substitute back into the quadratic formula
Now, substitute back into the formula:
\[ x = \frac{-1 \pm \sqrt{4a^2 + 12a + 9}}{2} \]

### Step 6: Factor the expression under the square root
Notice that \( 4a^2 + 12a + 9 \) can be factored:
\[ 4a^2 + 12a + 9 = (2a + 3)^2 \]
Thus, we can rewrite the equation as:
\[ x = \frac{-1 \pm (2a + 3)}{2} \]

### Step 7: Solve for the two possible values of \( x \)
Now, we can find the two roots:
1. For the positive case:
\[ x_1 = \frac{-1 + (2a + 3)}{2} = \frac{2a + 2}{2} = a + 1 \]
2. For the negative case:
\[ x_2 = \frac{-1 - (2a + 3)}{2} = \frac{-2a - 4}{2} = -a - 2 \]

### Final Result
The roots of the quadratic equation \( x^2 + x - (a+1)(a+2) = 0 \) are:
\[ x = a + 1 \quad \text{and} \quad x = -a - 2 \]

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