Question

If \(x^2 + 5ax + 6 = 0\) and \(x^2 + 3ax + 2 = 0\) have a common root, then that common root is:

• A. \(3 \quad \text{or} \quad -3\)
• B. \(2 \quad \text{or} \quad -2\)
• C. \(-2 \quad \text{or} \quad 3\)
• D. \(-3 \quad \text{or} \quad 2\)

Hint:

For quadratic equations with a common root, use substitution and elimination methods to simplify and solve for the root.

Answer 1

The Correct Option is B

We are given two quadratic equations: \( x^2 + 5ax + 6 = 0 \quad \text{and} \quad x^2 + 3ax + 2 = 0. \) Let the common root be \( r \). So, \( r \) satisfies both equations. 

Step 1: Substitute \( r \) in both equations: From \( x^2 + 5ax + 6 = 0 \), we have: \( r^2 + 5ar + 6 = 0 \quad \text{(1)} \) From \( x^2 + 3ax + 2 = 0 \), we have: \( r^2 + 3ar + 2 = 0 \quad \text{(2)}. \) 

Step 2: Subtract equation (2) from equation (1): \( (r^2 + 5ar + 6) - (r^2 + 3ar + 2) = 0 \) This simplifies to: \( 2ar + 4 = 0. \) Thus, we have: \( 2ar = -4 \quad \Rightarrow \quad ar = -2. \quad \cdots (3) \) 

Step 3: Now, substitute \( ar = -2 \) into equation (2): \( r^2 + 3ar + 2 = 0. \) Substitute \( ar = -2 \): \( r^2 + 3(-2) + 2 = 0 \quad \Rightarrow \quad r^2 - 6 + 2 = 0 \quad \Rightarrow \quad r^2 - 4 = 0. \) This simplifies to: \( r^2 = 4 \quad \Rightarrow \quad r = 2 \quad \text{or} \quad r = -2. \)