Question

The roots of the equation \( x^2 + 5x + 6 = 0 \) are:

• A. \( -2 \) and \( -3 \)
• B. \( 1 \) and \( 6 \)
• C. \( -1 \) and \( -6 \)
• D. \( -3 \) and \( -2 \)

Hint:

Factor quadratic equations whenever possible to find the roots quickly.

Answer 1

The Correct Option is A

Step 1: Write the equation
\[ x^2 + 5x + 6 = 0 \]
Step 2: Factorize the quadratic
We need two numbers that:
- Multiply to give the constant term (6)
- Add to give the coefficient of \( x \) (5)
Test pairs for 6:
- \( 1 \times 6 = 6 \), \( 1 + 6 = 7 \) (no)
- \( 2 \times 3 = 6 \), \( 2 + 3 = 5 \) (yes)
So: \[ x^2 + 5x + 6 = (x + 2)(x + 3) \] Set each factor to zero: \[ x + 2 = 0 \implies x = -2 \] \[ x + 3 = 0 \implies x = -3 \] Roots are \( -2 \) and \( -3 \).

Step 3: Verify with quadratic formula
For \( ax^2 + bx + c = 0 \): \[ x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \] Here, \( a = 1 \), \( b = 5 \), \( c = 6 \): \[ x = \frac{-5 \pm \sqrt{5^2 - 4 \cdot 1 \cdot 6}}{2 \cdot 1} \] \[ = \frac{-5 \pm \sqrt{25 - 24}}{2} = \frac{-5 \pm \sqrt{1}}{2} = \frac{-5 \pm 1}{2} \] \[ x = \frac{-5 + 1}{2} = \frac{-4}{2} = -2 \] \[ x = \frac{-5 - 1}{2} = \frac{-6}{2} = -3 \] Roots are \( -2 \) and \( -3 \).

Step 4: Check options
Option (1) \( -2 \) and \( -3 \) matches.