Question

If the equation \(x^2 + px + q = 0\) and \(x^2 + qx + p = 0\) have a common root then \(p + q + 1\) is equal to

• A. 0
• B. 1
• C. 2
• D. -1

Hint:

If two quadratics have a common root, subtract to find the root.

Answer 1

The Correct Option is A

To solve the problem, we begin by understanding that if two quadratic equations \(x^2 + px + q = 0\) and \(x^2 + qx + p = 0\) have a common root, then this root must satisfy both equations.

1. Let \(\alpha\) be the common root. Then, substituting \(\alpha\) into both equations, we get two conditions:
   • \(\alpha^2 + p\alpha + q = 0\)
   • \(\alpha^2 + q\alpha + p = 0\)
2. Subtracting the second equation from the first gives: \((\alpha^2 + p\alpha + q) - (\alpha^2 + q\alpha + p) = 0\) 
   Simplifying, we get: \(p\alpha + q - q\alpha - p = 0\) 
   which can be further simplified to: \((p-q)\alpha + q-p = 0\)
3. Simplifying further, we factor out common terms: \((p-q)(\alpha - 1) = 0\) 
    
4. This equation implies either:
   • \(p = q\), or
   • \(\alpha = 1\)
5. If \(\alpha = 1\), substitute \(\alpha = 1\) in either equation. Using the first equation, we have: 
   \(1^2 + p\cdot 1 + q = 0\) 
   Simplifying, we find: \(1 + p + q = 0\) 
   This leads to: \(p + q + 1 = 0\)
6. Therefore, the correct answer is 0.

This detailed analysis shows that if \(\alpha = 1\) is the common root, then \(p + q + 1 = 0\).